How do you simplify $({(8^{- \frac{1}{3}})}^{2})$ $((8^(-1/3))^2)$ ?

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How do you simplify $({(8^{- \frac{1}{3}})}^{2})$ $((8^(-1/3))^2)$ ?

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Answer 1 · 2016-05-22T19:38:10

$\frac{1}{4}$ $1/4$ $= 0.25$ $=0.25$

Explanation:

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

we know,
$8^{- \frac{1}{3}} = \frac{1}{2}$ $8^(-1/3) = 1/2$ by applying exponent rule i.e $a^{- b} = \frac{1}{a^{b}}$ $a^-b =1/(a^b$ $= \frac{1}{8^{\frac{1}{3}}}$ $=1/8^(1/3)$ and $8^{\frac{1}{3}} = 2$ $8^(1/3) = 2$

So,
${(\frac{1}{2})}^{2}$ $(1/2)^2$

$\frac{1}{4}$ $1/4$ = $0.25$ $0.25$

Answer 2 · 2016-05-22T20:35:32

$\frac{1}{4}$ $1/4$

Very slightly different approach!

Explanation:

Write as: $8^{- \frac{1}{3}} \times 8^{- \frac{1}{3}} = 8^{- \frac{2}{3}} = \frac{1}{\sqrt[3]{8^{2}}} = \frac{1}{4}$ $" "8^(-1/3)xx8^(-1/3) = 8^(-2/3) = 1/(root(3)(8^2)) = 1/4$

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