Prove $(\sin x)^{1/3}$ is an odd function

Question

A question asks to prove that $f(x)=(\sin x)^{1/3}$ is an odd function. I have begun by using the idea that for a function to be odd, $f(x)=-f(-x)$, such that... $$-f(-x)=-[\sin(-x)]^{1/3}$$ $$=-[-\sin x]^{1/3}$$ $$=-(-1)^{1/3}(\sin x)^{1/3}$$ From here though, I am unclear as to how I should proceed. Is there a basic algebraic error in my working to this point, or is there some simple step from here I am missing?

Answer 1

All that remains to do is to compute the real cube root of $-1$, which is $-1$ itself, and negate it to get 1, i.e. $-(-1)^{1/3}=1$. This proves $f(x)=-f(-x)$.

Answer 2

$(-1)^{1/3}=e^{\pi i/3+2k/3\pi i}$ for$k\in \mathbb{Z}$, so over the real nubmers, $(-1)^{1/3}=x$ has real solution $x=-1$. Nowe, $-f(-x)=--\sin(x)^{1/3}=f(x)$.

Answer 3

Alternatively you could argue that both $\sin(x)$ and $x^{1/3}$ are odd and that odd functions are closed under composition.