Q. 8.63 - Intermediate Algebra

Question

Simplify:

(a) $-{16}^{\frac{3}{2}}$

(b) $-{16}^{-\frac{3}{2}}$

(c) ${(-16)}^{-\frac{3}{2}}$

Step-by-Step Solution

Answer

The simplification of $-{16}^{\frac{3}{2}}$ is $-64$

The simplification of $-{16}^{-\frac{3}{2}}$ is $-\frac{1}{64}$

The simplification of ${(-16)}^{-\frac{3}{2}}$ has no real root.

Part (a) Step 1. Given

An expression $-{16}^{\frac{3}{2}}$

To simplify the given expression

Part (a) Step 2. Rewrite in radical form and simplify

Rewrite in radical form and simplify.

$-{16}^{\frac{3}{2}}=-{\left(\sqrt{16}\right)}^3$

          $=-{\left(\sqrt{4^2}\right)}^3$

          $=-\left(4^3\right)$

          $=-64$

Part (b) Step 1. Given

An expression $-{16}^{-\frac{3}{2}}$

To simplify the given expression

Part (b) Step 2. Rewrite in radical form and simplify

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

$-{16}^{-\frac{3}{2}}=-\left(\frac{1}{{16}^{{\displaystyle \frac{3}{2}}}}\right)$

Rewrite in radical form,

            $=-\left(\frac{1}{{\left(\sqrt{16}\right)}^3}\right)$

            $=-\left(\frac{1}{{\left(\sqrt{4^2}\right)}^3}\right)$

            $=-\left(\frac{1}{4^3}\right)$

            $=-\frac{1}{64}$

Part (c) Step 1. Given

An expression ${(-16)}^{-\frac{3}{2}}$

To simplify the given expression

Part (c) Step 2. Rewrite in radical form and simplify

Rewrite in radical form and simplify
${(-16)}^{-\frac{3}{2}}=\frac{1}{{(-16)}^{{\displaystyle \frac{3}{2}}}}$

               $=\frac{1}{{\left(\sqrt{-16}\right)}^3}$

There is no real number whose square root is $-16$.

So it has no real root.