How do you solve ${(\frac{1}{1000})}^{- \frac{1}{3}}$ $(1/1000)^(-1/3)$ ?

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Answer 1 · 2018-03-30T11:57:27

$10$ $10$

${(\frac{1}{1000})}^{- (\frac{1}{3})}$ $(1/1000)^-(1/3)$

$= \frac{1}{1000^{- (\frac{1}{3})}}$ $=1/1000^-(1/3)$

$= 1000^{\frac{1}{3}}$ $=1000^(1/3)$

$= \sqrt[3]{1000}$ $=root(3) 1000$

$= 10$ $=10$

Answer 2 · 2018-03-30T12:00:03

${(\frac{1}{1000})}^{- \frac{1}{3}} = 10$ $(1/1000) ^(-1/3) = 10$

${(\frac{1}{1000})}^{- \frac{1}{3}}$ $(1/1000) ^(-1/3)$

$= {({(\frac{1000}{1})}^{- 1})}^{- \frac{1}{3}}$ $= ((1000/1)^-1)^(-1/3)$

$= 1000^{- 1 \times - \frac{1}{3}}$ $= 1000^(-1 xx-1/3)$

$= 1000^{\frac{1}{3}}$ $= 1000^(1/3)$

$= {(10^{3})}^{\frac{1}{3}}$ $= (10^3)^(1/3)$

$= 10^{3 \times \frac{1}{3}}$ $= 10^(3xx1/3)$

$= 10$ $= 10$

How do you solve (11000)−13(1/1000)^(-1/3)?