Let $x$ be a negative real number and $0<\alpha <1$, I would like to know why ${(x)}^{\alpha }$ is not a negative real number.
For example wolfram alpha shows that $$ {(-5)}^{0.33 }=0.856097+1.47919i $$ but it's not $-1.70906$.
My question:
When is it true that
$$ {(x)}^{\alpha }=-{(x)}^{\alpha}, $$ for $x<0$ and $0<\alpha <1$ ?
$(x)^\alpha=-(x)^\alpha$ does not make sense since this implies that $x^\alpha=0$. You are probably asking if $(x)^\alpha=-(-x)^\alpha$ is true in general for $x<0$ and $0<\alpha<1$.
The answer is NO.
For $x<0$ and $0<\alpha<1$, one has to refers to complex analysis in order to make sense of $x^\alpha$. If $\alpha=\frac{n}{m}$ for some integers $m$ and $n$, then essentially you need to know the $m$-th root of unity. If otherwise $\alpha$ is irrational, then $$ x^\alpha:=e^{\alpha\log x} $$ and the right hand side is a long story.
[Added:] $x^\alpha=-(-x)^\alpha$ for $x<0$ and $0<\alpha<1$ could be true if one stricts oneself to the real setting and further assumes that $\alpha=\dfrac{1}{m}$ where $m$ is an odd interger. For instance, you do have $$ (-2)^{1/3}=-(2)^{1/3} $$ where $y=(-2)^{1/3}$ is defined as a real number such that $y^3=-2$.