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# Understanding nth Roots

Suppose we know that ${a}^{3}=8$. We want to find what number raised to the 3rd power is equal to 8. Since ${2}^{3}=8$, we say that 2 is the cube root of 8. The nth root of $a$ is a number that, when raised to the nth power, gives $a$. For example, $-3$ is the 5th root of $-243$ because ${\left(-3\right)}^{5}=-243$. If $a$ is a real number with at least one nth root, then the principal nth root of $a$ is the number with the same sign as $a$ that, when raised to the nth power, equals $a$. The principal nth root of $a$ is written as $\sqrt[n]{a}$, where $n$ is a positive integer greater than or equal to 2. In the radical expression, $n$ is called the index of the radical.

### A General Note: Principal nth Root

If $a$ is a real number with at least one nth root, then the principal nth root of $a$, written as $\sqrt[n]{a}$, is the number with the same sign as $a$ that, when raised to the nth power, equals $a$. The index of the radical is $n$.

### Example 10: Simplifying nth Roots

Simplify each of the following:
1. $\sqrt[5]{-32}$
2. $\sqrt[4]{4}\cdot \sqrt[4]{1,024}$
3. $-\sqrt[3]{\frac{8{x}^{6}}{125}}$
4. $8\sqrt[4]{3}-\sqrt[4]{48}$

### Solution

1. $\sqrt[5]{-32}=-2$ because ${\left(-2\right)}^{5}=-32 \\ \text{ }$
2. First, express the product as a single radical expression. $\sqrt[4]{4,096}=8$ because ${8}^{4}=4,096 \\$
3. $\begin{array}{cc}\\ \frac{-\sqrt[3]{8{x}^{6}}}{\sqrt[3]{125}}\hfill & \text{Write as quotient of two radical expressions}.\hfill \\ \frac{-2{x}^{2}}{5}\hfill & \text{Simplify}.\hfill \\ \end{array}$
4. $\begin{array}{cc}\\ 8\sqrt[4]{3}-2\sqrt[4]{3}\hfill & \text{Simplify to get equal radicands}.\hfill \\ 6\sqrt[4]{3} \hfill & \text{Add}.\hfill \\ \end{array}$

### Try It 10

Simplify.
1. $\sqrt[3]{-216}$
2. $\frac{3\sqrt[4]{80}}{\sqrt[4]{5}}$
3. $6\sqrt[3]{9,000}+7\sqrt[3]{576}$
Solution