If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$ 
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$ 
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$

I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
  b^2 &= 4 - a^2\\
  b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\begin{align*}
  (a + ib)^3 &= 8\\
  a + \sqrt{4-a^2} &= 2\\
  \sqrt{4-a^2} &= 2 - a\\
  2 - a &= 2 - a
\end{align*}
$$
Therefore if $(a + ib)^3 = 8$, then $a^2 + b^2 = 4$.

$$a^3+3a^2(ib)-3ab^2+(ib)^3=8$$
Equating the real & the imaginary parts,
$$a^3-3ab^2=8;3a^2b-b^3=0\iff b(3a^2-b^2)=0$$
Case $\#1:b=0\implies a^3=8\implies a=2$
Case $\#2:3a^2-b^2=0\iff b^2=3a^2$ 
and we have $a(a^2-3b^2)=8\implies a[a^2-3(3a^2)]=8\iff a^3=-1\implies a=-1$ as $a$ is real
$\implies b^2=?$
