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 $a^2 + b^2 = 4$.

Hint: $$
a^2 + b^2 = (a+ib)(a-ib)
$$

as $(a+ib)^3 = 8$, $(a-ib)^3 = 8$ and 
$$
(a^2 + b^2)^3 = 8^2 = 64 \implies a^2 + b^2 = 4
$$
