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7.1: Defining nth Roots

Difficulty Level: At Grade Created by: CK-12
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The volume of a cube is found to be \begin{align*}343s^7\end{align*}343s7. What is the length of each side of the cube?

nth Roots

So far, we have seen exponents with integers and the square root. In this concept, we will link roots and exponents. First, let’s define additional roots. Just like the square and the square root are inverses of each other, the inverse of a cube is the cubed root. The inverse of the fourth power is the fourth root.

\begin{align*}\sqrt[3]{27}=\sqrt[3]{3^3}=3, \sqrt[5]{32}=\sqrt[5]{2^5}=2\end{align*}273=333=3,325=255=2

The \begin{align*}n^{th}\end{align*}nth root of a number, \begin{align*}x^n\end{align*}xn, is \begin{align*}x, \sqrt[n]{x^n}=x\end{align*}x,xnn=x. And, just like simplifying square roots, we can simplify \begin{align*}n^{th}\end{align*}nth roots.

Let's find \begin{align*}\sqrt [6]{729}\end{align*}7296.

To simplify a number to the sixth root, there must be 6 of the same factor to pull out of the root.

\begin{align*}729=3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3=3^6\end{align*}729=333333=36

Therefore, \begin{align*}\sqrt[6]{729}=\sqrt[6]{3^6}=3\end{align*}7296=366=3. The sixth root and the sixth power cancel each other out. We say that 3 is the sixth root of 729.

From this problem, we can see that it does not matter where the exponent is placed, it will always cancel out with the root.

\begin{align*}\sqrt[6]{3^6}&=\sqrt[6]{3}^6 \ or \ \left(\sqrt[6]{3} \right)^6\\ \sqrt[6]{729}&=\left(1.2009 \ldots\right)^6\\ 3&=3\end{align*}36672963=366 or (36)6=(1.2009)6=3

So, it does not matter if you evaluate the root first or the exponent.

The \begin{align*}n^{th}\end{align*}nth Root Theorem: For any real number \begin{align*}a\end{align*}a, root \begin{align*}n\end{align*}n, and exponent \begin{align*}m\end{align*}m, the following is always true: \begin{align*}\sqrt[n]{a^m}=\sqrt[n]{a}^m=\left(\sqrt[n]{a}\right)^m\end{align*}amn=anm=(an)m.

Now, let's evaluate the following expressions without a calculator.

  1. \begin{align*}\sqrt[5]{32^3}\end{align*}3235

If you solve this problem as written, you would first find \begin{align*}32^3\end{align*}323 and then apply the \begin{align*}5^{th}\end{align*}5th root.


However, this would be very difficult to do without a calculator. This is an example where it would be easier to apply the root and then the exponent. Let’s rewrite the expression and solve.


  1. \begin{align*}\sqrt{16}^3\end{align*}163

This problem does not need to be rewritten. \begin{align*}\sqrt{16}=4\end{align*}16=4 and then \begin{align*}4^3 = 64\end{align*}43=64.

Finally, let's simplify the following.

  1. \begin{align*}\sqrt[4]{64}\end{align*}644

To simplify the fourth root of a number, there must be 4 of the same factor to pull it out of the root. Let’s write the prime factorization of 64 and simplify.

\begin{align*}\sqrt[4]{64}=\sqrt[{\color{red}4}]{{\color{red}2 \cdot 2\cdot 2\cdot 2}\cdot 2\cdot 2}=2\sqrt[4]{4}\end{align*}644=2222224=244

Notice that there are 6 2’s in 64. We can pull out 4 of them and 2 2’s are left under the radical.

  1. \begin{align*}\sqrt[3]{\frac{54x^3}{125y^5}}\end{align*}54x3125y53

Just like simplifying fractions with square roots, we can separate the numerator and denominator.

\begin{align*}\sqrt[3]{\frac{{54x^3}}{{125y^5}}}= \frac{\sqrt[3]{54x^3}}{\sqrt[3]{125y^5}}=\frac{\sqrt[{\color{red}3}]{2 \cdot{\color{red}3\cdot 3\cdot 3}\cdot {\color{red}x^3}}}{\sqrt[{\color{blue}3}]{{\color{blue}5\cdot 5\cdot 5\cdot y^3}\cdot y^2}}=\frac{3x\sqrt[3]{2}}{5y\sqrt[3]{y^2}}\end{align*}54x3125y53=54x33125y53=2333x33555y3y23=3x235yy23

Notice that because the \begin{align*}x\end{align*}x is cubed, the cube and cubed root cancel each other out. With the \begin{align*}y\end{align*}y-term, there were five, so three cancel out with the root, but two are still left under radical.


Example 1

Earlier, you were asked to find the length of each side of the cube. 

Recall that the volume of a cube is \begin{align*}V = s^3\end{align*}V=s3, where s is the length of each side. So to find the side length, take the cube root of \begin{align*}343z^7\end{align*}343z7.

First, you can separate this number into two different roots, \begin{align*}\sqrt[3]{343} \cdot \sqrt[3]{z^7}\end{align*}3433z73. Now, simplify each root.

\begin{align*}\sqrt[3]{343}\cdot \sqrt[3]{z^7} = \sqrt[3]{7^3}\cdot \sqrt[3]{z^3\cdot z^3 \cdot z}= 7z^2 \sqrt[3]{z}\end{align*}3433z73=733z3z3z3=7z2z3

Therefore, the length of the cube's side is \begin{align*}7z^2 \sqrt[3]{z}\end{align*}7z2z3.

Simplify each expression below, without a calculator.

Example 2


First, you can separate this number into two different roots, \begin{align*}\sqrt[4]{625} \cdot \sqrt[4]{z^8}\end{align*}6254z84. Now, simplify each root.

\begin{align*}\sqrt[4]{625}\cdot \sqrt[4]{z^8} = \sqrt[4]{5^4}\cdot \sqrt[4]{z^4\cdot z^4}= 5z^2\end{align*}6254z84=544z4z44=5z2

When looking at the \begin{align*}z^8\end{align*}z8, think about how many \begin{align*}z^4\end{align*}z4 you can even pull out of the fourth root. The answer is 2, or a \begin{align*}z^2\end{align*}z2, outside of the radical.

Example 3


\begin{align*}32 = 2^5\end{align*}32=25, which means there are not 7 2's that can be pulled out of the radical. Same with the \begin{align*}x^5\end{align*}x5 and the \begin{align*}y\end{align*}y. Therefore, you cannot simplify the expression any further.

Example 4


Write out 9216 in the prime factorization and place factors into groups of 5.

\begin{align*}\sqrt[5]{9216}&=\sqrt[5]{\boxed{2 \cdot 2\cdot 2\cdot 2\cdot 2}\cdot \boxed{2\cdot 2 \cdot 2\cdot 2\cdot 2} \cdot 3\cdot 3}\\ &=\sqrt[5]{2^5\cdot 2^5 \cdot 3^2}\\ &=2\cdot 2 \sqrt[5]{3^2}\\ &=4\sqrt[5]{9}\end{align*}92165=2222222222335=2525325=22325=495

Example 5


Reduce the fraction, separate the numerator and denominator and simplify.

\begin{align*}\sqrt[3]{\frac{40}{175}}=\sqrt[3]{\frac{8}{35}}=\frac{\sqrt[3]{2^3}}{\sqrt[3]{35}}={\color{red}{\frac{2}{\sqrt[3]{35}}}\cdot} {\color{red}{\frac{\sqrt[3]{35^2}}{\sqrt[3]{35^2}}}}=\frac{2\sqrt[3]{1225}}{35}\end{align*}

In the red step, we rationalized the denominator by multiplying the top and bottom by \begin{align*}\sqrt[3]{35^2}\end{align*}, so that the denominator would be \begin{align*}\sqrt[3]{35^3}\end{align*} or just 35. Be careful when rationalizing the denominator with higher roots!


Reduce the following radical expressions.

  1. \begin{align*}\sqrt[3]{81}\end{align*}
  2. \begin{align*}\sqrt[4]{625}^3\end{align*}
  3. \begin{align*}\sqrt{9^5}\end{align*}
  4. \begin{align*}\sqrt[5]{128}\end{align*}
  5. \begin{align*}\sqrt{\sqrt{10000}}\end{align*}
  6. \begin{align*}\sqrt{\frac{25}{8}}^4\end{align*}
  7. \begin{align*}\sqrt[6]{64}^5\end{align*}
  8. \begin{align*}\sqrt[3]{\frac{8}{81}}^2\end{align*}
  9. \begin{align*}\sqrt[4]{\frac{243}{16}}\end{align*}
  10. \begin{align*}\sqrt[3]{24x^5}\end{align*}
  11. \begin{align*}\sqrt[4]{48x^7y^{13}}\end{align*}
  12. \begin{align*}\sqrt[5]{\frac{160x^8}{y^7}}\end{align*}
  13. \begin{align*}\sqrt[3]{1000x^6}^2\end{align*}
  14. \begin{align*}\sqrt[4]{\frac{162x^5}{y^3z^{10}}}\end{align*}
  15. \begin{align*}\sqrt{40x^3y^4}^3\end{align*}

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 7.1. 

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n^{th} root

The n^{th} root of a number, x^n, is x. Therefore: \sqrt[n]{x^n}=x.

nth root

The n^{th} root of a number, x^n, is x. Therefore: \sqrt[n]{x^n}=x.

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Difficulty Level:
At Grade

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Date Created:
Mar 12, 2013
Last Modified:
Sep 07, 2016
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