If $a^{-1}$ has a cube root, so does $a$. 
If $a^{-1}$ has a cube root [in a group], then so does $a$.

Attempt
Suppose $a^{-1} = bbb.$ Then, $b^{-1}(a^{-1})b^{-1} = b^{-1}(bbb)b^{-1} \to b = b^{-1}(a^{-1})b^{-1}.$ Now I am not sure if it holds up. I have to think about it a bit more.
 A: First of all, you should specify that $a$ is an element of an arbitrary group (I assume that's what you're interested in).
Your proof only works for elements $a$ such that $a^{-1}=aaa$, since you assume that right from the start for no reason. In other words, you're assuming that not only does $a^{-1}$ have a cube root, but that that cube root is also its inverse. What you have after that is a correct proof that such elements do indeed have a cube root (namely $a^{-1}$).
What you perhaps meant to write was "Suppose there exists $b$ such that $a^{-1}=bbb$", in other words, assume $a^{-1}$ has a cube root, and then figure out what it is. With this correction, the rest of your proof falls apart immediately: multiplying by $a^{-1}$ we just get $a^{-1}a^{-1}a^{-1}=a^{-1}bbba^{-1}$, which tells us nothing about $a$ having an inverse.
Thus, your proof should start with "Let $a^{-1}=bbb$". Note that it's let and not suppose. You're not supposing anything, you're just giving a name to the cube root, which was already supposed to exist in the problem statement.
A: Hint If $a^{-1} = bbb$ then 
$$(a^{-1})^{-1}=???$$
A: Note that $(bbb)^{-1}=b^{-1}b^{-1}b^{-1}$
So, from $a^{-1}=bbb$ you have:
$$a=\left(a^{-1}\right)^{-1}=(bbb)^{-1}=b^{-1}b^{-1}b^{-1}
$$
