How do you simplify ${(- 8)}^{- \frac{1}{3}}$ $(-8)^(-1/3)$ ?

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How do you simplify ${(- 8)}^{- \frac{1}{3}}$ $(-8)^(-1/3)$ ?

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Answer 1 · 2018-05-03T09:59:08

$- \frac{1}{2}$ $-1/2$

Explanation:

$a^{- n} = \frac{1}{a^{n}}$ $a^-n=1/a^n " "$ [this is a proved result]

So,

$- 8^{- \frac{1}{3}} = {(- \frac{1}{8})}^{\frac{1}{3}}$ $-8^(-1/3)=(-1/8)^(1/3)$

${(- \frac{1}{8})}^{\frac{1}{3}} = \sqrt[3]{- \frac{1}{8}} = - \frac{1}{2}$ $(-1/8)^(1/3)= root(3)(-1/8)=-1/2$

Answer 2 · 2018-05-03T10:01:39

${(- 8)}^{- \frac{1}{3}} = - \frac{1}{2}$ $(-8)^(-1/3)=-1/2$

Explanation:

${(- 8)}^{- \frac{1}{3}} =^{-3} \sqrt{- 8}$ $(-8)^(-1/3)=^"-3"sqrt(-8)$
$= \frac{1}{^3 \sqrt{- 8}}$ $=1/(^"3"sqrt(-8))$
$= - \frac{1}{2}$ $=-1/2$
\0/ here's our answer !

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How do you simplify ${(- 8)}^{- \frac{1}{3}}$ $(-8)^(-1/3)$ ?

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