Question

Simplify:

(a)${27}^{\frac{2}{3}}$

(b)${81}^{-\frac{3}{2}}$

(c)${16}^{-\frac{3}{4}}$

Short answer

The simplification of ${27}^{\frac{2}{3}}$is$9$

The simplification of ${81}^{-\frac{3}{2}}$is$\frac{1}{729}$

The simplification of ${16}^{-\frac{3}{4}}$is$\frac{1}{8}$

Step-by-step solution

Part (a) Step 1. Given

An expression ${27}^{\frac{2}{3}}$

To simplify the given expression

Part (a) Step 2. Change to radical form

The power of the radical is the numerator of the exponent, $2$.
The index of the radical is the denominator of the exponent, $3$.

${27}^{\frac{2}{3}}={\left(\sqrt[3]{27}\right)}^2$

$={\left(\sqrt[3]{3^3}\right)}^2$

$=3^2$

$=9$

Part (b) Step 1. Given

An expression${81}^{-\frac{3}{2}}$

To simplify the given expression

Part (b) Step 2. Change to radical form

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

${81}^{\frac{-3}{2}}=\frac{1}{{81}^{{\displaystyle \frac{3}{2}}}}$

The power of the radical is the numerator of the exponent,$3$.
The index of the radical is the denominator of the exponent, $2$.

${81}^{-\frac{3}{2}}=\frac{1}{{81}^{{\displaystyle \frac{3}{2}}}}$

$=\frac{1}{{\left(\sqrt{9^2}\right)}^3}$

$=\frac{1}{9^3}$

$=\frac{1}{729}$

Part (c) Step 1. Given

An expression${16}^{-\frac{3}{4}}$

To simplify the given expression

Part (c) Step 2. Change to radical form

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

${16}^{-\frac{3}{4}}=\frac{1}{{16}^{{\displaystyle \frac{3}{4}}}}$

The power of the radical is the numerator of the exponent, $3$.
The index of the radical is the denominator of the exponent, $4$.

${16}^{-\frac{3}{4}}=\frac{1}{{\left(\sqrt[4]{16}\right)}^3}$

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

$=\frac{1}{2^3}$

$=\frac{1}{8}$