Roots Explained Completely: Square Roots, Cube Roots, All Tricks, and Methods

Roots are one of the most important ideas in mathematics. They are the reverse of powers, and they help us understand what number was multiplied by itself to produce a given result. A square root asks, “What number squared gives this number?” A cube root asks, “What number cubed gives this number?” A fourth root, fifth root, and higher roots ask the same question with a higher power.

Roots show up in geometry, algebra, measurement, equations, physics, and number patterns. They also appear in competitive exams, where speed and accuracy matter. Many students learn only the basic symbol and a few perfect squares, but real root mastery goes much deeper. You need to know how to estimate, simplify, compare, extract roots, use prime factorization, handle surds, and apply fast shortcuts.

This article explains roots clearly from the very beginning to advanced level. It covers square roots, cube roots, fourth roots, fifth roots, nth roots, root laws, shortcuts, and the fastest hacks used by strong math students.

What Is a Root?

A root is the reverse of a power. Yes, you have heard it right. Just like multiplication is reverse of divide. Subtraction is reverse of addition. The same way, roots work. If 3² = 9.
Then: √9 = 3. Because 3 is the number that was squared to make 9.
$\sqrt[2]{9} = 3. ~~~ reverse\ it ~~~ 3^2 = 9.$ Get it.

The simple formula you can use: $\sqrt[n]{a} = b ~~ means ~~ b^n = a.$ Use it every time to confirm your final answer is correct or incorrect.

Note: We normally do not write Square root 2. So if you do not see any number assume it is always Square root. $\sqrt9 = \sqrt[2]{9}$. Both are the same.

If: 2³ = 8.
Then: ³√8 = 2.
Because 2 is the number that was cubed to make 8.
Look: $2^3 = 8.$

So a root asks:
“What number, when raised to this power, gives the original number?”

Square Root

The square root of a number is the number that, when multiplied by itself, gives that number. Confusing! Don’t worry, look at the examples.
1st Example: √25 = 5
Because: 5² = 5 × 5 = 25.

2nd Example: √81 = 9
Because: 9² = 9 × 9 = 81

Square roots are the most common type of root. We will learn all the techniques of roots later, so stay attentive and learn your own pace.

Why It Is Called Square Root

A square has equal sides. If the area of a square is known, the side length is the square root of the area. In simple words, square is the multiplied value of its own value twice.
Example: If a square has area 49, then side = √49 = 7. Because 7 × 7 = 49. So roots are deeply connected to geometry.

More examples:

• $\sqrt{121} = 11 ~~~ reverse\ it ~~~ 11^2 = 121.$
• $\sqrt{196} = 14 ~~~ reverse\ it ~~~ 14^2 = 196.$
• $\sqrt{289} = 17 ~~~ reverse\ it ~~~ 17^2 = 289.$
• $\sqrt{529} = 23 ~~~ reverse\ it ~~~ 23^2 = 529.$
• $\sqrt{841} = 29 ~~~ reverse\ it ~~~ 29^2 = 841.$

Cube Root

The cube root of a number is the number that, when multiplied by itself three times, gives that number. Haha, in simple words, a number is multiplied by the same number three times.
1st Example: ³√27 = 3
Because: 3³ = 3 × 3 × 3 = 27. Like we have learnt already roots reverse is exponent / power.

2nd Example: ³√64 = 4
Because: 4³= 4 × 4 × 4 = 64.

More examples:

• $\sqrt{729} = 9 ~~~ reverse\ it ~~~ 9^3 = 729.$
• $\sqrt{2197} = 13 ~~~ reverse\ it ~~~ 13^3 = 2197.$
• $\sqrt{35,937} = 33 ~~~ reverse\ it ~~~ 33^3 = 35,937.$

Cube roots are very useful in volume problems because volume often comes from multiplying length × width × height.

Fourth Root, Fifth Root, and Higher Roots

Roots can have any positive whole number as the index. Increasing the index numbers won’t change the process. The whole process still stays the same like above in square and cube.

Examples:
⁴√16 = 2 because 2⁴ = 16
⁵√32 = 2 because 2⁵ = 32
⁶√64 = 2 because 2⁶ = 64.

The small number written near the root sign is called the index. If no index is written, the root is usually a square root.
Example: √49 means ²√49. The 2 is index understood.

Perfect Squares, Perfect Cubes, and Perfect Powers

A perfect square is a number that is the square of a whole number.
Examples: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

A perfect cube is a number that is the cube of a whole number.
Examples: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.

A perfect fourth power, fifth power, and so on are numbers formed by raising whole numbers to those powers.
Examples:
2⁴ = 16
3⁴ = 81
2⁵ = 32
3⁵ = 243

Knowing perfect powers is the foundation of fast root work.

Root and Exponent Connection

Roots and exponents are reverse operations. We have discussed it above and shown you many examples. But if you don’t know exponent / power yet. Learn from here: Click.

The Formula: $a^{\frac{m}{n}} = (\sqrt[n]{81})^m$

If: a² = b
Then: √b = a.
Get it. That is how you change index on the root and exponent on the number.
If: a³ = b
Then: ³√b = a.
This connection helps you move between root form and exponent form. You can increase the index to 4, 5, 6, 7, 8, to any number. The reverse exponent will have the same number as well. But what about a fraction as an exponent.

Fraction Exponent Examples:

• $16^{\frac12} = \sqrt{16} = 4.$
• $27^{\frac13} = \sqrt[3]{27} = 3.$
• $81^{\frac14} = \sqrt[4]{81} = 3.$
• $3,125^{\frac15} = \sqrt[5]{3,125} = 5.$
• $46,656^{\frac23} = (\sqrt[3]{46,656})^2 = 6.$
• $1,048,576^{\frac45} = (\sqrt[5]{1,048,576})^4 = 2.$

The Fastest Trick is Memorize: Perfect Method

The genius pros in calculation all do that. They remember most used numbers in all calculation to save a lot of their time when solving problems. You should do the same if you are trying to solve problems quickly.

Square Root by Memorization of Perfect Squares

The fastest way to handle many square root questions is to memorize perfect squares up to at least 30² or 50². Here are key ones:

• 1² = 1
• 2² = 4
• 3² = 9
• 4² = 16
• 5² = 25
• 6² = 36
• 7² = 49
• 8² = 64
• 9² = 81
• 10² = 100
• 11² = 121
• 12² = 144
• 13² = 169
• 14² = 196
• 15² = 225
• 16² = 256
• 17² = 289
• 18² = 324
• 19² = 361
• 20² = 400
• 21² = 441
• 22² = 484
• 23² = 529
• 24² = 576
• 25² = 625
• 26² = 676
• 27² = 729
• 28² = 784
• 29² = 841
• 30² = 900

This is one of the most powerful memory tools in school math.

Cube Root by Memorization of Perfect Cubes

For cube roots, memorize common cubes.

• 1³ = 1
• 2³ = 8
• 3³ = 27
• 4³ = 64
• 5³ = 125
• 6³ = 216
• 7³ = 343
• 8³ = 512
• 9³ = 729
• 10³ = 1000
• 11³ = 1331
• 12³ = 1728
• 13³ = 2197
• 14³ = 2744
• 15³ = 3375
• 16³ = 4096
• 17³ = 4913
• 18³ = 5832
• 19³ = 6859
• 20³ = 8000

These numbers help with fast recognition and estimation.

The Laws of Roots

Roots follow some useful laws. You have learnt some above but there are more. The formula, the steps, and the process is part of the law. Law meaning here, you cannot break it.

Product Rule for Same Index

When the index number is the same in all values, we can apply this shortcut.
Rule: $\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}$
$\sqrt[n]{a} \times \sqrt[n]{b} \times \sqrt[n]{c} = \sqrt[n]{abc}$
It keeps going on like this.

Examples:

• $\sqrt[2]{2} \times \sqrt[2]{8} = \sqrt[2]{2\times8} = \sqrt[2]{16} = 4$
• $\sqrt[3]{4} \times \sqrt[3]{2} = \sqrt[3]{4\times2} = \sqrt[3]{8} = 2$
• $\sqrt[4]{9,261} \times \sqrt[4]{21} = \sqrt[4]{9261\times21} = \sqrt[4]{194,481} = 21$
• $\sqrt[2]{50} \times \sqrt[2]{6} \times \sqrt[2]{3} = \sqrt[2]{50\times6\times3} = \sqrt[2]{900} = 30$
• $\sqrt[3]{48} \times \sqrt[3]{24} \times \sqrt[3]{12} = \sqrt[3]{48\times24\times12} = \sqrt[3]{13,824} = 24$

Opposite is the Smartest Hack

What it means by opposite is that when the index of root is the same in all multiplying roots’ value then we merge them and put the index on the root of the total like above, right? What if we use the same method but do it opposite to make things easier for us.

For Examples:

• $\sqrt[2]{11,025} = \sqrt[2]{49} \times \sqrt[2]{25} \times \sqrt[2]{9} = 7\times5\times3 = 105$.
• $\sqrt[3]{512} = \sqrt[3]{64} \times \sqrt[3]{8} = 4\times2 = 8$.

Once you know multiplication then you can easily separate them into two, three or even more numbers. But there are a lot of other methods out there which are far easier then these, we will learn them all after this.

Quotient Rule for Same Index

We have learnt how multiplication of root values work. Now, let’s learn how division of root values work. The process is almost same but with a few distinct steps.
Rule: $\sqrt[n]{a} \div \sqrt[n]{b} = \sqrt[n]{\frac{a}{b}}$. The only issue is that we can do this with only two values at a time. If there are more then, we do it again and again like this.

$\sqrt[n]{a} \div \sqrt[n]{b} \div \sqrt[n]{c} = \sqrt[n]{\frac{a}{b}} \div \sqrt[n]{c}$. After dividing a and b, suppose we get A then we do $\sqrt[n]{A} \div \sqrt[n]{c} = \sqrt[n]{\frac{A}{c}} = \sqrt[n]{D}$. The final root value we get after their division as well. Learn from examples below.

Examples:

• $\sqrt{49} \div \sqrt7 = \sqrt{\frac{49}{7}} = \sqrt7$
  
• $\sqrt[2]{24} \div \sqrt[2]{3} \div \sqrt[2]{4} \div \sqrt[2]{2}$
  $\\ \sqrt[2]{\frac{24}{3}} \div \sqrt[2]{4} \div \sqrt[2]{2} \\ \sqrt[2]{8} \div \sqrt[2]{4} \div \sqrt[2]{2} \\ \sqrt[2]{\frac{8}{4}} \div \sqrt[2]{2} \\ \sqrt[2]{2} \div \sqrt[2]{2} \\ \sqrt[2]{\frac{2}{2}} = \sqrt[2]{1} = 1$

See, this is how we do divide, try more examples like these. Never do mistake like dividing third and fourth root together. Always divide in the sequence first and second then the result of it by third root, after it with fourth and so on.

Power Inside a Root

We have already talked about it that a power is reverse of a root. So technically if the power is inside a root that mostly makes it easy to solve unless both power and index are different.
Rule: $\sqrt[n]{a^n} = a$.
When both n are the same, they become null. But if both are different then, we solve it by expanding.
$\sqrt[n]{a^m}$

When Power and Index are the same

The solution of it is the easiest and fastest.
Examples:

• $\sqrt[2]{5^2} = 5$
• $\sqrt[3]{7^3} = 7$
• $\sqrt[5]{9^5} = 9$
• Be careful with negative values and even roots in advanced settings. So you should expand and understand it very well. Look carefully.
• $\sqrt[2]{-3^2} = \sqrt[2]{-3\times-3} = \sqrt[2]{3^2}= 3$. Get it how negative becomes positive because of exponent. You may know the rules, right. Negative × Negative = Plus. What about Positive × Negative = Negative.
  So:
• $\sqrt[3]{-4^3} = \sqrt[3]{-4\times-4\times-4} = \sqrt[3]{4^2\times-4}$
  $= \sqrt[3]{-4^3}= -4$. I hope you are getting it all.
  
• $\sqrt[4]{-5^4} = \sqrt[4]{-5\times-5\times-5\times-5}$
  $= \sqrt[4]{5^2\times-5\times-5} = \sqrt[4]{-5^3\times-5}$
  $= \sqrt[4]{5^4}= 5$. If you are getting hard time to get it, talk to our experts. It is completely free. They will help you learning anything.
  
• $\sqrt[5]{-12^5} = \sqrt[5]{-12\times-12\times-12\times-12\times-12}$
  $= \sqrt[5]{12^2\times-12\times-12\times-12} = \sqrt[5]{-12^3\times-12\times-12}$
  $\sqrt[5]{12^4\times-12} = \sqrt[5]{-12^5}= -12$

When power and index are different

It becomes slightly difficult and complex when both are different. There are two ways to solve it. One is simple by solving both power and index one by one. Second is by expanding power like above we saw.

Examples:

1. $\sqrt[2]{3^3} = \sqrt[2]{3^2\times3} = \sqrt[2]{3^2} \times \sqrt[2]{3}$
   $= 3 \times \sqrt[2]{3}= 5.196…$
   What if it is a negative value then it will go on like above, the same way.
   $\sqrt[2]{-3^3} = \sqrt[2]{3^2\times-3} = \sqrt[2]{3^2} \times \sqrt[2]{-3}$
   $= 3 \times \sqrt[2]{-3}= -5.196…$
   
2. $\sqrt[4]{4^6}$ The rule says: $\sqrt[n]{4^{m}} = 4^{\frac{m}{n}}$. We must always Simplify: $\frac64 = \frac32$.
   So, $\sqrt[2]{4^3} = \sqrt[2]{4 \times4 \times 4} = \sqrt[2]{4^2 \times 4}$
   $= 4 \times \sqrt[2]{4} = 8.$ Ask our experts if you still have any doubts.
   
3. $\sqrt[3]{9^2}$ Always simplify: 9 = 3² . So why not we replace it.
   $\sqrt[3]{9^2} = \sqrt[3]{(3^2)^2} = \sqrt[3]{3^4}$
   $\sqrt[3]{3^4} = \sqrt[3]{3 \times3 \times 3 \times 3} = \sqrt[3]{3^3 \times 3}$
   $= 3 \times \sqrt[3]{3} = 4.3267$
   
4. There is another trick to simply solve any problems like these is to do this way.
   $\sqrt[6]{12^5}$. The trick: $\sqrt[n]{12^m} = 12^{\frac{m}{n}}$
   Rewrite the exponent: $\frac{m}{n} = \frac56 = 1\frac{-1}{6}$
   Do you remember mixed fraction from fractions topic?
   $12^{1\frac{-1}{6}} = 12^1 \times 12^{-\frac{1}{6}} = \frac{12}{\sqrt[6]{12}}$.
   Now we solve it.
   $\frac{12}{1.513} = 7.931…$ approximately
   
5. The same process when exponent is higher than index.
   $\sqrt[4]{10^7}$. The trick: $\sqrt[n]{10^m} = 10^{\frac{m}{n}}$
   Rewrite the exponent: $\frac{m}{n} = \frac74 = 2\frac{-1}{4} ~~ or ~~ 1\frac{3}{4}$
   That means, the many mixed fraction we have, the many ways we can solve. You have seen the negative above, let’s pick positive one, okay.
   $10^{1\frac{3}{4}} = 10^1 \times 10^{\frac{3}{4}} = 10 \times \sqrt[4]{10^3}$.
   Now we solve it. The exponent 3 of 10 means total 3 zeros.
   $10 \times \sqrt[4]{1000} = 10 \times 5.62341 = 56.2341…$ approximately
   Haha, complex, right. I showed you to understand the concepts behind it. But in real exam, you should solve problems like these directly.
   Meaning: exponent 7 on 10 denotes total 7 zeros.
   $\sqrt[4]{10000000} = 56.234…$

Root of a Root

Like the name suggest root of a root means we multiply or divide the root based on problem.
Rule: $\sqrt[n]{\sqrt[m]{a}} = \sqrt[n×m]{a} = \sqrt[nm]{a}$
If we need to smaller it then: $\sqrt[mn]{a} = \sqrt[m×n]{a} = \sqrt[m]{\sqrt[n]{a}}$

Examples:

• $\sqrt[3]{\sqrt[4]{6}} = \sqrt[3×4]{6} = \sqrt[12]{6}$
• $\sqrt[5]{\sqrt[2]{9}} = \sqrt[5×2]{9} = \sqrt[10]{9}$
• $\sqrt[14]{10} = \sqrt[7×2]{10} = \sqrt[7]{\sqrt[2]{10}}$
• $\sqrt[14]{10} = \sqrt[7×2]{10} = \sqrt[7]{\sqrt[2]{10}}$

This is more advanced but useful in algebra.

Rationalizing the Denominator

In some expressions, roots appear in the denominator. In basic algebra, we often remove roots from the denominator by multiplying top and bottom with the same root. It is a method or a way to solve it. Sometimes in examination, this way of solution is asked, so use then.
Note: It doesn’t mean this is the only way to solve any root in denominator. You can use any method, unless the question asked you to solve it this way.

Examples:

• $\frac{1}{\sqrt{2}}$
  Multiply top and bottom by √2:
  $\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
  When completely same roots multiplied together, we keep one product and remove the root like above.
  
• $\frac{3}{\sqrt{5}}$
  Multiply top and bottom by √5:
  $\frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$.

This is called rationalizing.

Comparing Roots

Sometimes in examination, question asks to compare roots quickly. How to compare them, which is bigger than which. Let’s learn:

Root comparison: The larger number has the larger root.
Example:
√49 < √64 because 49 < 64

For approximate comparison: When we compare solutions of a root.
√10 is more than 3 but less than 4
Since:
3² = 9
4² = 16.
Get it. Rather than solving roots, you can guess which one is bigger or smaller mostly. When you see a problem asking to compare, use these approach.

How to Solve Roots: All Methods and Tricks

There are several methods to solve roots. The best one depends on the number. But we will learn all methods here whether to solve square root, cube root or even bigger ones. We will solve and show your visuals solutions. Because so far you have learnt tricks and exponent to directly solve a root, but we will learn the visual methods where without an exponent you can solve it from a scratch like a calculator does.

Product Rule: Expanding the Number

Product is the number that is inside the root, what we usually do is to simplify it by expanding or rewriting it. Suppose if a product is 36, we can rewrite it like “6 × 6” or “4 × 9” or “3 × 12” or “2 × 18.” By rewriting them this way, we can easily do root of these numbers. Experts always use techniques like these to solve big values by making them smaller and solve. The best part the answer will be the same in all rewrites.
Examples:

• 64 and its rewrites:
  $\sqrt{64} = 8$
  $\sqrt{8 × 8} = \sqrt8 \times \sqrt8 = 2.828… \times 2.828… = 8$
  $\sqrt{32 × 2} = \sqrt{32} \times \sqrt2 = 5.656… \times 1.414… = 8$
  $\sqrt{16 × 4} = \sqrt{16} \times \sqrt4 = 4 \times 2 = 8$. This rewrite is the easiest one so far. The experts they know these multiples and they take advantage of it. I just showed you all ways but you just need to filter the easiest one always remember number 16 and 4 and other numbers which roots perfectly and in your rewrite attempt to bring them.
• 36 = 9 × 4. Let’s check.
  $\sqrt{36} = 6$
  $\sqrt{9 × 4} = \sqrt{9} \times \sqrt4 = 3 \times 2 = 6$.

Perfect Solution Recognition

We have talked about it above the Perfect Square and Perfect Cube method. You may have seen many examples above. If you remember all and more then it will make calculation easy.
Examples:

• √144 = 12 Because: 12² = 144
• ³√125 = 5 Because: 5³ = 125
• ³√1000 = 10 Because: 10³ = 1000

This is the fastest method when the number is perfection.

Prime Factorization

The method is almost same as LCM with a distinct new feature. We factorize the product number and then pick common factors based on index number. If index number is square (2) then we pair the factors of two. If the index is cube (3), we pair the factors of three. Based on index, the pair is decided.
Examples:

1. $\sqrt[]{144}$
   Prime Factors: Square Root
   $\begin{array}{|c|l|r|} \hline Calculations & Factors & Products \\ \hline 144÷2 = 72 & 2 & \bold{144} \\ \hline 72÷2 = 36 & 2 & 72 \\ \hline 36÷2 = 18 & 2 & 36 \\ \hline 18÷2 = 9 & 2 & 18 \\ \hline 9÷3 = 3 & 3 & 9 \\ \hline 3÷3 = 1 & 3 & 3 \\ \hline \ \bold{Pair\ of\ 2} & ~ & 1 \\ \hline \end{array}$
   $\sqrt[]{144} = \sqrt[]{(2×2)(2×2)(3×3)}$
   $= \sqrt[]{(2^2)(2^2)(3^2)} = 2×2×3 = 12.$
   This is how we do prime factors.
   
2. $\sqrt[]{3600}$
   Prime Factors: Square Root
   $\begin{array}{|c|l|r|} \hline Calculations & Factors & Products \\ \hline 3600÷2 = 1800 & 2 & \bold{3600} \\ \hline 1800÷2 = 900 & 2 & 1800 \\ \hline 900÷2 = 450 & 2 & 900 \\ \hline 450÷2 = 225 & 2 & 450 \\ \hline 225÷3 = 75 & 3 & 225 \\ \hline 75÷3 = 25 & 3 & 75 \\ \hline 25÷5 = 5 & 5 & 25 \\ \hline 5÷5 = 1 & 5 & 5 \\ \hline \ \bold{Pair\ of\ 2} & ~ & 1 \\ \hline \end{array}$
   $\sqrt[]{3600} = \sqrt[]{(2×2)(2×2)(3×3)(5×5)}$
   $= \sqrt[]{(2^2)(2^2)(3^2)(5^2)} = 2×2×3×5 = 60.$
   Prime factorization can also be used if the number is not a perfect square.
   
3. $\sqrt[]{1768}$
   Prime Factors: Square Root
   $\begin{array}{|c|l|r|} \hline Calculations & Factors & Products \\ \hline 1768÷2 = 884 & 2 & \bold{1768} \\ \hline 884÷2 = 442 & 2 & 884 \\ \hline 442÷2 = 221 & 2 & 442 \\ \hline 221÷13 = 17 & 13 & 221 \\ \hline 17÷17 = 1 & 17 & 17 \\ \hline \ \bold{Pair\ of\ 2} & ~ & 1 \\ \hline \end{array}$
   $\sqrt[]{1768} = \sqrt[]{(2×2)×2×13×17}$
   $= \sqrt[]{(2^2)(2)(13)(17)} = 2\sqrt[]{2×13×17} = 42.047…$ Approximately
   When we do not have a pair of the factors we must solve them by root.
   
4. $\sqrt[3]{432}$
   Prime Factors: Cube Root
   $\begin{array}{|c|l|r|} \hline Calculations & Factors & Products \\ \hline 432÷2 = 216 & 2 & \bold{432} \\ \hline 216÷2 = 108 & 2 & 216 \\ \hline 108÷2 = 54 & 2 & 108 \\ \hline 54÷2 = 27 & 2 & 54 \\ \hline 27÷3 = 9 & 3 & 27 \\ \hline 9÷3 = 3 & 3 & 9 \\ \hline 3÷3 = 1 & 3 & 3 \\ \hline \ \bold{Pair\ of\ 3} & ~ & 1 \\ \hline \end{array}$
   $\sqrt[3]{216} = \sqrt[3]{(2×2×2)×2×(3×3×3)}$
   $= \sqrt[3]{(2^3)(3^3)×2} = 2×3\sqrt[3]{2}$
   $= 6\sqrt[3]{2} = 7.5594…$
   When we use index as cube then we organize pair of three.
   
5. $\sqrt[5]{6048}$
   Prime Factors: Fifth Root
   $\begin{array}{|c|l|r|} \hline Calculations & Factors & Products \\ \hline 6048÷2 = 3024 & 2 & \bold{6048} \\ \hline 3024÷2 = 1512 & 2 & 3024 \\ \hline 1512÷2 = 756 & 2 & 1512 \\ \hline 756÷2 = 378 & 2 & 756 \\ \hline 378÷2 = 189 & 2 & 378 \\ \hline 189÷3 = 63 & 3 & 189 \\ \hline 63÷3 = 21 & 3 & 63 \\ \hline 21÷3 = 7 & 3 & 21 \\ \hline 7÷7 = 1 & 7 & 7 \\ \hline \ \bold{Pair\ of\ 5} & ~ & 1 \\ \hline \end{array}$
   $\sqrt[5]{6048} = \sqrt[5]{(2×2×2×2×2)(3×3×3)×7}$
   $= \sqrt[5]{(2^5)(3^3)×7} = 2×\sqrt[5]{3^3×7} = 5.705…$
   When we use index as fifth then we organize pair of five.

This is the simplified radical form. It is a simple and faster method, to solve problem till the point of radical form but when it is about getting full final answer then it is not the right choice.

Long Division Method for Roots

This is the standard manual method for finding roots digit by digit. Once you learn this method, you can solve any root problems, it is kind of like a division but we can say Power Division.
The Rule: Exponent the number with the index always.

Examples:

1. $\sqrt[]{1521} = 39$
   Long Division: Square Root
   Since it is Square root, so we must follow these patterns.
   Dividend: In Pairs of Two: 15 | 21
   Divisor: Must be powered by Two because of Square root.
   New Divisor Formula: $\bold{(Place\ Value\ Quotient × Index + x) × x}$
   Place Value Quotient: It goes like Tens, Hundreds, Thousands, Ten-Thousands, Hundred-Thousands, Millions. In simple words, it shifts one position higher each time. It would be better to just look visually how to do.
   Look PVQ:
   3 (It is Ones, so after ones comes Tens: 30 × 2)
   60
   +9
   69 (69 × 9 = 621)
   $\begin{array}{|c|l|r|} \hline Divisor & Dividend & Quotient \\ \hline \bold{3^{2}} & \bold{1521} & \bold{39} \\ \hline (3×3) & -9 \ \ \ \ \ \ & (\bold{3}) \\ \hline \hline (30 × 2 + 9) × 9 \\ \bold{69} & ~~~6\bold{21} & ~ \\ \hline (69×9) & -621\ & (3\bold{9}) \\ \hline \ Remainder & \bold{0} & ~ \\ \hline \end{array}$
   This is how we do Square root of Long division. You can also apply this formula “$\bold{(20q + x) × x}$ ” here but I want you to learn the concept why and how the formula made this way.
   
2. $\sqrt[]{1894} = 43.52011029…$ approximately
   Long Division: Square Root
   Another Square root, but a little more complex.
   Dividend: In Pairs of Two: 18 | 94
   Divisor: Must be powered by Two because of Square root.
   New Divisor Formula: $\bold{(20q + x) × x}$
   q stands for: Current Quotient
   Decimal / Point: In division, normally we use point only single time and it always brings One Zero in every attempt. But in roots, based on index, we bring Two Zeros.
   $\begin{array}{|c|l|r|} \hline Divisor & Dividend & Quotient \\ \hline \bold{4^{2}} & \bold{1894} & \bold{43.5201102} \\ \hline (4×4) & -16 ~~~~~~~~~ & (\bold{4}) \\ \hline \hline (20{×}4 + 3) × 3 \\ \bold{83} & ~~~2\bold{94} & ~ \\ \hline (83×3) & -249\ & (4\bold{3}) \\ \hline Take\ Point & ~~~~~45 & (43\bold{.}) \\ \hline \hline (20{×}43 + 5) × 5 \\ \bold{865} & ~~~~~~~~~~~45\bold{00} & ~ \\ \hline (865×5) & ~~~~~~ -4325\ & (43.\bold{5}) \\ \hline \hline (20{×}435 + 2) × 2 \\ \bold{8702} & ~~~~~~~~~~~~~~~~~~175\bold{00} & ~ \\ \hline (8702×2) & ~~~~~~~~~~~~~ -17404\ & (43.5\bold{2}) \\ \hline Take\ extra\ Zero \\ Point\ brings\ 2\ zeros & ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~96\bold{00} & (43.52\bold{0}) \\ \hline \hline (20{×}43520 × 2 + 1) × 1 \\ \bold{870401} & ~~~9600\bold{00} & ~ \\ \hline (870401×1) & -870401 & (43.520\bold{1}) \\ \hline \hline (20{×}435201 + 1) × 1 \\ \bold{8704021} & ~~~89599\bold{00} & ~ \\ \hline (8704021×1) & -8704021\ & (43.5201\bold{1}) \\ \hline Take\ extra\ Zero\ again \\ Point\ brings\ 2\ zeros & ~~~~~~~~~~~255879\bold{00} & (43.52011\bold{0}) \\ \hline \hline (20{×}43520110 + 2) × 2 \\ \bold{870402202} & ~~~25587900\bold{00} & ~ \\ \hline (870402202×2) & -1740804404 & (43.250110\bold{2}) \\ \hline It\ keeps\ going\ on\ like & this,\ so\ let’s\ stop. & ~ \\ \hline \ Remainder & \bold{817,985,596} & ~ \\ \hline \end{array}$
   The purpose to solve this long is to teach you and guide you how to do it. We do not want you to get scared of it. The truth is after point, “Solve Max three digits” that is it. Like 43.250… That is enough.
   
3. There is a simple and fastest trick, just for Square root. Instead of using the New Division Formula, use this method.
   $~~~4 \\ +\bold{4} \\ ~~~~8\bold{3} \\ ~+3 \\ ~~~~86\bold{5} \\ ~~~~+5 \\ ~~~~870\bold{2} \\ ~~~~~~+2 \\ ~~~~8704\bold{0} \\ ~~~~87040\bold{1} \\ ~~~~~~~~~~~+1 \\ ~~~~870402\bold{1} \\ ~~~~~~~~~~~~~~+1 \\ ~~~~8704022\bold{0} \\ ~~870402202$
   The whole New Divisor goes like this, isn’t it far simpler than the formula, but it only works in square roots.
   
4. $\sqrt[3]{17,576} = 26$
   Long Division: Cube Root
   Since it is a Cube root, so we must follow these patterns.
   Dividend: In Pairs of Three: 17 | 576
   Divisor: Must be powered by Three because of Square root.
   New Divisor Formula: $\bold{(300q^2 + 30qx + x^2) × x}$
   Why x is 6: The complex part is to solve this formula every time, just to find the largest value against 9576. Try using 5 and 7 as x. You will know why 6!
   $\begin{array}{|c|l|r|} \hline Divisor & Dividend & Quotient \\ \hline \bold{2^{3}} & \bold{17,576} & \bold{26} \\ \hline (2×2×2) & -8~~~~~~~~~~ & (\bold{2}) \\ \hline \hline 300q^2 = 300{×}2^2 = 1200 \\ 30qx = 30{×}2{×}6 = +360 \\ ~~~~~~~~~~~~~~~~~~~~x^2 = 6^2 = +36 \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\bold{1596} & ~~9\bold{576} & ~ \\ \hline (1596×6) & -9576\ & (2\bold{6}) \\ \hline \ Remainder & \bold{0} & ~ \\ \hline \end{array}$
   This is how we do Cube root of Long division.
   
5. The whole method and process are the same in all Long Division Method of Roots but the only change is in the New Divisor Formula. Let’s check more formula.
   $\sqrt[4]{a}$ when Index is 4 in root: $\bold{(4000q^3+ 600q^2x + 40qx^2 + x^3) × x}$
   $\sqrt[5]{a}$ index is 5 now: $\bold{(50000q^4 + 10000q^3x+ 1000q^2x^2 + 50qx^3 + x^4) × x}$
   
6. A common useful approximation of square root is:
   √2 ≈ 1.414
   √3 ≈ 1.732
   √5 ≈ 2.236
   √10 ≈ 3.162

These values are very useful in exams. I know the method seems difficult but it is for knowledge purpose. In exams mostly you can give answers in Radical form using Prime factorization like above.

Estimation Method

As the name suggest we shorten the root’s product value by separating them into smaller more Perfect version then solve them easily, so two things are most important. One, shorten to Perfect root level. Two, memorize some most used square and cube roots. I have given you those above, right?
The Formula for square: $\sqrt{a+b} = \sqrt{a} + \frac{b}{2\sqrt{a}}$
The formula for cube: $\sqrt[3]{a+b} = \sqrt[3]{a} + \frac{b}{3(\sqrt[3]{a})^2}$
The formula for 4th: $\sqrt[4]{a+b} = \sqrt[4]{a} + \frac{b}{4(\sqrt[4]{a})^3}$
Note: The result of the formula is not mostly accurate but it always gives the closest result.

Examples:

1. $\sqrt{50} = 7.07106…$
   Shorten to Perfect Root: $\sqrt{49+1}$
   $\sqrt{49} = 7, because\ 7^2 = 49.$
   Apply Formula: $\sqrt{a} + \frac{b}{2\sqrt{a}}$
   $\sqrt{49} + \frac{1}{2\sqrt{49}} \\ 7 + \frac{1}{14} = 7.07142…$
   The correct answer is 7.071… approx.
   
2. Let’s try a cube:
   $\sqrt[3]{75} = 4.2172…$
   Shorten to Perfect Root: $\sqrt[3]{64+11}$
   $\sqrt[3]{64} = 4, because\ 4^3 = 64.$
   Apply Formula: $\sqrt[3]{a} + \frac{b}{3(\sqrt[3]{a})^2}$
   $\sqrt[3]{64} + \frac{11}{3(\sqrt[3]{64})^2} \\ 4 + \frac{11}{3\times16} = 4.22916…$
   The correct answer is 4.2… approx.
   
3. Let’s try a 4th:
   $\sqrt[4]{164} = 3.5802…$
   Shorten to Perfect Root: $\sqrt[4]{81+83}$
   $\sqrt[4]{81} = 3, because\ 3^4 = 81.$
   Apply Formula: $\sqrt[4]{a} + \frac{b}{4(\sqrt[4]{a})^3}$
   $\sqrt[4]{81} + \frac{83}{4(\sqrt[4]{81})^3} \\ 3 + \frac{83}{4\times27} = 3.76851…$
   The correct answer is 3.6… approx.

That is why the method is called estimation, because we do not get accurate answer but we are mostly close in approximate direction. This method is fastest if you need to solve the root completely and options are given. You can easily pick the most closed option this way.

Learn All Fastest Trick and Hack Against Root Used by Genius

The pros, the genius, and all masters use these tricks to solve root problems by saving a lot of their time. You should learn some too, so you can save time in examination.

Fast Trick for Square Roots of Numbers Ending in 25

Numbers ending in 25 often have beautiful square root patterns. Looks a little rough at first but it becomes easy.
Rule for Digits end with 5: Suitable Number × Next Number | 25

Examples:

• $\sqrt{125} = 11.1803…$ approx.
  Why because: Suitable Number × Next Number | 25
  But to get 1 we: 1 × Next Number? Impossible, next number should be 2, but then we cannot get 1 by multiplying it to 2. The trick doesn’t work here.
  
• $\sqrt{225} = 2 ~|~ 25$
  To find 2, we: 1 × 2 = 2, so 1 and 5.
  From 25, we ignore the 2 and only pick 5.
  Correct answer: 15.
  Must verify: $15^2 = 225.$
  
• $\sqrt{425} = 20.6155$. The trick won’t work. Because we cannot find the value to put in the formula.
  4 | 25 = 2 × 3? We cannot use the same number, because we need Next Number.
  Trick won’t work here.
  
• $\sqrt{625} = 6 ~|~ 25$
  To find 6, we: 2 × 3 = 6, so 2 and 5.
  From 25, we ignore the 2 and only pick 5.
  Correct answer: 25.
  Must verify: $25^2 = 625.$
  
• $\sqrt{1225} = 12 ~|~ 25$
  To find 12, we: 3 × 4 = 12, so 3 and 5.
  From 25, we ignore the 2 and only pick 5.
  Correct answer: 35.
  Verify: $35^2 = 1225.$
  
• $\sqrt{2025} = 20 ~|~ 25$
  To find 20, we: 4 × 5 = 20, so 4 and 5.
  From 25, we ignore the 2 and only pick 5.
  Correct answer: 45.
  Verify: $45^2 = 2025.$
  
• $\sqrt{3025} = 30 ~|~ 25$
  To find 30, we: 5 × 6 = 30, so 5 and 5.
  From 25, we ignore the 2 and only pick 5.
  Correct answer: 55.
  Verify: $55^2 = 3025.$
  
• $\sqrt{5625} = 56 ~|~ 25$
  To find 56, we: 7 × 8 = 56, so 7 and 5.
  From 25, we ignore the 2 and only pick 5.
  Correct answer: 75.
  Verify: $75^2 = 5625.$
  
• $\sqrt{46225} = 462~|~ 25$
  To find 462, we: 21 × 22 = 462, so 21 and 5.
  From 25, we ignore the 2 and only pick 5.
  Correct answer: 215.
  Verify: $215^2 = 46225.$

This is a very powerful shortcut. As long as you can find numbers that fits the formula, you are good to go. I know it only works with Perfect Root Solutions but still quite useful.

Root Simplification by Breaking into Factors

The best trick to solve any root problem in no and very short time is to make factors and leave non-perfect root as it is, meaning, do not solve it completely. Only solve perfect root completely when you create factors.
Surd: A surd is a root that cannot be simplified to a rational number. Like these √2, √3, √5, and ³√2. These are irrational numbers and are very common in geometry and algebra. Look below examples and see unsolved roots. They are surds.

Examples:

• √200
  Factors: 200 = 100 × 2
  So: √200 = √100 × √2 = 10√2.
  
• √72
  Factors: 72 = 36 × 2
  So: √72 = √36 × √2 = 6√2.
  
• √12
  Factors: 12 = 4 × 3
  So: √12 = √4 × √3 = 2√3.
  
• ³√54
  Factors: 54 = 27 × 2
  So: ³√54 = ³√27 × ³√2 = $3\sqrt[3]{2}$.

This is one of the most important simplification methods. No need to solve the whole, use prime factors and perfect powers.

Fourth Roots, Fifth Roots, and Higher Roots

For higher roots, the same idea applies. Use prime factors and perfect powers for them as well, but they are rather hard to solve this is why the best approach is using exponent as a guide.

Examples:

• ⁴√81 = 3
  Because: 3⁴ = 81.
  
• $\sqrt[4]{648}$
  Factors: 648 = 81 × 8
  So: $\sqrt[4]{648} = \sqrt[4]{81} \times \sqrt[4]{8}= 3\sqrt[4]{8}$.
  
• ⁵√32 = 2
  Because: 2⁵ = 32.
• ⁶√64 = 2
  Because: 2⁶ = 64.

The best way to handle higher roots is to recognize perfect powers and prime factorize. But what if it is imperfect (not perfect) then use simplification approach from above.

Conclusion

Roots are one of the most powerful ideas in mathematics because they reverse powers and reveal the hidden number behind an expression. Square roots, cube roots, and higher roots all follow the same core idea: find the number that raises to a given power to make the original value. Once you understand perfect powers, prime factorization, estimation, digit grouping, and simplification methods, roots become much easier and much faster.

Fast Exam Hacks for Roots

1. Memorize perfect squares and cubes.
2. Break numbers into known factors.
3. Use prime factorization to simplify radicals.
4. Estimate between nearby perfect powers.
5. Use long division for exact square roots.
6. Recognize ending patterns like 25, 00, and 625.
7. Convert roots to exponents when algebra becomes easier.
8. For cube roots, remember digit grouping and perfect cubes.

The strongest students do not just memorize a few square roots. They learn the structure behind roots, recognize patterns, and use the right method for each problem. That is how root questions become quick, clean, and confident.