How do you simplify (root4(x^3)*root4(x^5))^(-2)?

Aug 28, 2017

See a solution process below:

Explanation:

First, we can use this rule to combine the radicals within the parenthesis:

$\sqrt[n]{\textcolor{red}{a}} \cdot \sqrt[n]{\textcolor{b l u e}{b}} = \sqrt[n]{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}}$

${\left(\sqrt[4]{\textcolor{red}{{x}^{3}}} \cdot \sqrt[4]{\textcolor{b l u e}{{x}^{5}}}\right)}^{-} 2 = {\left(\sqrt[4]{\textcolor{red}{{x}^{3}} \cdot \textcolor{b l u e}{{x}^{5}}}\right)}^{-} 2$

Next, use this rule for exponents to combine the terms within the radical:

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

${\left(\sqrt[4]{\textcolor{red}{{x}^{3}} \cdot \textcolor{b l u e}{{x}^{5}}}\right)}^{-} 2 = {\left(\sqrt[4]{{x}^{\textcolor{red}{3} + \textcolor{b l u e}{5}}}\right)}^{-} 2 = {\left(\sqrt[4]{{x}^{8}}\right)}^{-} 2$

Then, we can use this rule to rewrite the radical into an exponent:

$\sqrt[\textcolor{red}{n}]{x} = {x}^{\frac{1}{\textcolor{red}{n}}}$

${\left(\sqrt[\textcolor{red}{4}]{{x}^{8}}\right)}^{-} 2 = {\left({\left({x}^{8}\right)}^{\frac{1}{\textcolor{red}{4}}}\right)}^{-} 2$

Next, we can use this rule to simplify the inner exponents:

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

${\left({\left({x}^{\textcolor{red}{8}}\right)}^{\textcolor{b l u e}{\frac{1}{4}}}\right)}^{-} 2 = {\left({x}^{\textcolor{red}{8} \times \textcolor{b l u e}{\frac{1}{4}}}\right)}^{-} 2 = {\left({x}^{2}\right)}^{-} 2$

We can use the same rule to reduce the outer exponents:

${\left({x}^{\textcolor{red}{2}}\right)}^{\textcolor{b l u e}{- 2}} = {x}^{\textcolor{red}{2} \times \textcolor{b l u e}{- 2}} = {x}^{-} 4$

We can now use this rule to eliminate the negative exponent:

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

${x}^{\textcolor{red}{- 4}} = \frac{1}{x} ^ \textcolor{red}{- - 4} = \frac{1}{x} ^ 4$