Guidance for a complex number proof I am given $\left(x+iy\right)^{\frac{1}{3}}=a+ib$, and I need to prove $4\left(a^{2\ }-b^{2}\right)=\frac{x}{a}+\frac{y}{b}$.
The first "key" thing (I hope it's actually useful!) I notice is that we are only considering the real parts (I'm pretty sure, anyway).
I tried "cheating" but extending the LHS and RHS to $4\left(a-b\right)\left(a+b\right)=\frac{xb+ya}{ab}$ but I don't think that's particularly helpful for me because I don't immediately know what to do.
I also tried rewriting the given pieces of information: $\left(a+ib\right)^{3}$ to find
$x=a^{3}-3ab^{2}$
$y=3a^{2}b-b^{3}$
Right now my plan in to get everything on LHS then RHS in terms of $x$ and $y$ to show they are equal, but I am having trouble with that. If you have a better method as well, feel free to comment.

You already (correctly) figured out that $x+iy = (a+ib)^3$ implies
$$
 x = a(a^2-3b^2) \, ,\\
 y = b(3a^2 -b^2) \, .
$$
It follows that
$$
\frac x a + \frac y b = (a^2-3b^2)+ (3a^2 -b^2) = 4(a^2 -b^2)
$$
Another way is to compute
$$
 (x+iy)(a+ib) = (a+ib)^4 = (a^4-6a^2b^2+b^4) + i(4a^3 b-4ab^3)
$$
and compare the imaginary parts:
$$
 xb + ya = 4a^3 b-4ab^3 = 4ab (a^2-b^2) \, .
$$
If $ab \ne 0$ then the desired formula follows.
