Question

For which non-zero real values of `x` is `-x^-5 = (-x)^-5` ?

Answer 1

All `x!=0 in RR`.

Explanation:

We have:

`-1/(x)^5=1/((-x)^5)`.

Observe that for every value of `x!=0` in `x^5`, if `x` is negative, then `x^5` is negative; the same is true if `x` is positive: `x^5` will be positive.

Therefore we know that in our equality, if `x<0`,

`-1/(x)^5=1/((-x)^5) rArr -1/(-x)^5=1/((-(-x))^5)`,

and from what we previously observed,

`-1/(-x)^5=1/((-(-x))^5) rArr 1/x^5=1/x^5`.

The same is true if `x>0`,

`-1/(x)^5=1/((-x)^5) rArr -1/x^5=-1/x^5`.

Therefore this equality is true for all `x!=0 in RR`.