<s>[INST] 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 $a^2 + b^2 = 4$.
 [/INST] We have $a+ib=2w$ where $w^3=1$
Taking modulus on either sides $\sqrt{a^2+b^2}=2|w|$

Directly taking  modulus on either sides on $(a+ib)^3=8,$
we get $(\sqrt{a^2+b^2})^3=8\implies (a^2+b^2)^3=8^2=(2^3)^2=2^6$
As $a^2+b^2>0, a^2+b^2=\sqrt[3]{2^6}=(2^6)^{\frac13}=?$
</s>