How do you simplify $16^{- \frac{2}{3}}$ $16^(-2/3)$ ?

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Answer 1 · 2015-03-30T14:59:20

$16^{- \frac{2}{3}}$ $16^(-2/3)$

= $\frac{1}{16^{\frac{2}{3}}}$ $1/(16^(2/3))$

$= \frac{1}{\frac{{(16^{2})}^{1}}{3}}$ $= 1/((16^2)^1/3)$

$= \sqrt[3]{4 \times 4 \times 4 \times 4}$ $= root(3)(4xx4xx4xx4)$

$= \frac{1}{4 \sqrt[3]{4}}$ $=1/(4 root(3)(4))$

$= \frac{{(\sqrt[3]{4})}^{2}}{4 {(\sqrt[3]{4})}^{3}}$ $= ((root(3)(4))^2)/(4 (root(3)(4))^3)$

$= \frac{\sqrt[3]{4^{2}}}{16}$ $= (root(3)(4^2))/16$

$= \frac{2 \sqrt{2}}{16}$ $= (2sqrt(2))/(16)$

$= \frac{\sqrt{2}}{8}$ $=sqrt(2)/8$

How do you simplify 16^(-2/3)16−2316^(-2/3)?