Zoe Oneal

## Answered question

2020-10-28

Solve the examples below. The answer should contain only positive indicators without fractional indicators in the denominator. Leave Rational (Not Radical) Answers:
$1\right)\left(343{b}^{3}{\right)}^{\frac{4}{3}}$
$2\right)\left(216{x}^{6}{\right)}^{-\frac{5}{3}}$
$3\right)3{a}^{-2}{b}^{\frac{5}{4}}×2{a}^{-\frac{7}{4}}{b}^{\frac{4}{3}}$

### Answer & Explanation

unett

Skilled2020-10-29Added 119 answers

Step 1
$\left(343{b}^{3}{\right)}^{\frac{4}{3}}=\left(\left(7b{\right)}^{3}{\right)}^{\frac{4}{3}}$
$=\left(7b{\right)}^{3×\frac{4}{3}}$
$=\left(7b{\right)}^{4}$
$={7}^{4}{b}^{4}$
$=2401{b}^{4}$
Step 2
$\left(216{x}^{6}{\right)}^{-\frac{5}{3}}=\left({6}^{3}{x}^{6}{\right)}^{-\frac{5}{3}}$
$=\left({6}^{3}{\right)}^{-\frac{5}{3}}×\left({x}^{6}{\right)}^{-\frac{5}{3}}$
$={6}^{-3×-\frac{5}{3}}×{x}^{6×-\frac{5}{3}}$
$={6}^{-5}{x}^{-10}$
$=\frac{1}{{6}^{5}{x}^{10}}$
$=\frac{1}{\left(6{x}^{2}{\right)}^{5}}$
Step 3
$3{a}^{-2}{b}^{\frac{5}{4}}×2{a}^{-\frac{7}{4}}{b}^{\frac{4}{3}}=6{a}^{-2-\frac{7}{4}}{b}^{\frac{5}{4}+\frac{4}{3}}$
$=6{a}^{-\frac{15}{4}}{b}^{\frac{31}{12}}$
$=\frac{6{b}^{\frac{31}{12}}}{{a}^{\frac{15}{4}}}$
$=\frac{6{b}^{\frac{31}{12}}{a}^{\frac{1}{4}}}{{a}^{\frac{15}{4}+\frac{1}{4}}}$
$=\frac{6{b}^{\frac{31}{12}}{a}^{\frac{1}{4}}}{{a}^{\frac{16}{4}}}$
$=\frac{6{b}^{\frac{31}{12}}{a}^{\frac{1}{4}}}{{a}^{4}}$

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