For which non-zero real values of $x$ $x$ is $- x^{- 6} = {(- x)}^{- 6}$ $-x^(-6) = (-x)^(-6)$ ?

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For which non-zero real values of $x$ $x$ is $- x^{- 6} = {(- x)}^{- 6}$ $-x^(-6) = (-x)^(-6)$ ?

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Answer 1 · 2018-02-16T20:02:02

None

Explanation:

If $x \neq 0$ $x != 0$ then $x^{2} > 0$ $x^2 > 0$ and we find:

$- x^{- 6} = - \frac{1}{{(x^{2})}^{3}} < 0$ $-x^(-6) = -1/(x^2)^3 < 0$

${(- x)}^{- 6} = \frac{1}{{({(- x)}^{2})}^{3}} = \frac{1}{{(x^{2})}^{3}} > 0$ $(-x)^(-6) = 1/((-x)^2)^3 = 1/(x^2)^3 > 0$

So they cannot be equal.

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For which non-zero real values of $x$ $x$ is $- x^{- 6} = {(- x)}^{- 6}$ $-x^(-6) = (-x)^(-6)$ ?

1 Answer

Explanation:

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For which non-zero real values of $x$ $x$ is $- x^{- 6} = {(- x)}^{- 6}$ $-x^(-6) = (-x)^(-6)$ ?