Q 140E - Intermediate Algebra

Question

In the following exercises, simplify.

Part (a): ${\left(-32\right)}^{\frac{1}{5}}$

Part (b): ${\left(243\right)}^{-\frac{1}{5}}$

Part (c): $-{125}^{\frac{1}{3}}$

Step-by-Step Solution

Answer

Part (a): On simplifying, we know the solution is not real.

Part (b): On simplifying, we get $\frac{1}{3}$.
Part (c): On simplifying, we get $-5$.

Part (a) Step 1. Rewrite as a root.

Consider the given question,

${\left(-32\right)}^{\frac{1}{5}}$

Rewrite as a fifth root,

$=\sqrt[5]{\left(-32\right)}$

On simplifying the expression,

$=\sqrt[5]{{\left(-2\right)}^5}$

The solution is not real.

Part (b) Step 1. Rewrite as a root.

Consider the given question,

${\left(243\right)}^{-\frac{1}{5}}$

Rewrite using the property $a^{-n}=\frac{1}{a^n}$,

$=\frac{1}{{\left(243\right)}^{{\displaystyle \frac{1}{5}}}}$

Rewrite as a fifth root,

$=\frac{1}{\sqrt[5]{\left(243\right)}}$

On simplifying the expression,

$$\begin{gathered} =\frac{1}{\sqrt[5]{3^5}} \\ =\frac{1}{3} \end{gathered}$$

Part (c) Step 1. Rewrite as a root.

Consider the given question,

$-{125}^{\frac{1}{3}}$

The exponent only applies to the $125$.

Rewrite as a cube root,

$=-\sqrt[3]{125}$

On simplifying the expression,

$$\begin{gathered} =-\sqrt[3]{5^3} \\ =-5 \end{gathered}$$