# How do you evaluate and simplify (7^3/4^3)^(-1/3)?

Jun 20, 2017

See a solution process below:

#### Explanation:

First, use this rule of exponents to eliminate the outer exponent:

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left({7}^{\textcolor{red}{3}} / {4}^{\textcolor{red}{3}}\right)}^{\textcolor{b l u e}{- \frac{1}{3}}} = {7}^{\textcolor{red}{3} \times \textcolor{b l u e}{- \frac{1}{3}}} / {4}^{\textcolor{red}{3} \times \textcolor{b l u e}{- \frac{1}{3}}} \implies {7}^{-} \frac{1}{4} ^ - 1$

Next, use these rules of exponents to eliminate the negative exponents:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$ and $\frac{1}{x} ^ \textcolor{red}{a} = {x}^{\textcolor{red}{- a}}$

${7}^{-} \frac{1}{4} ^ - 1 \implies {7}^{\textcolor{red}{- 1}} \times \frac{1}{4} ^ \textcolor{red}{- 1} = \frac{1}{7} ^ \textcolor{red}{- - 1} \times {4}^{\textcolor{red}{- - 1}} = \frac{1}{7} ^ 1 \times {4}^{1} \implies {4}^{1} / {7}^{1}$

Now, use this rule of exponents to complete the simplification:

${a}^{\textcolor{red}{1}} = a$

${4}^{\textcolor{red}{1}} / {7}^{\textcolor{red}{1}} = \frac{4}{7}$

Jun 20, 2017

$\frac{4}{7}$

#### Explanation:

${\left({7}^{3} / {4}^{3}\right)}^{- \frac{1}{3}}$ is the same as ${\left({4}^{3} / {7}^{3}\right)}^{+ \frac{1}{3}}$

This is the same as $\frac{\sqrt[3]{{4}^{3}}}{\sqrt[3]{{7}^{3}}} \text{ "=" } \frac{4}{7}$