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

There have already been enough answers, but none addresses why your proposed answer is incorrect. Mine will do just that.
The most basic mistake is that you use what you want to prove as if it was true. This is a logical fallacy called circular reasoning. There is also a number of other mistakes:


*

*From $b^2=4-a^2$ it follows that $b= \pm \sqrt{4-a^2}$ and not necessarily that $b=\sqrt{4-a^2}$.

*Similarly, from $(a+ib)^3=8$ it doesn't follow that $a+ib = 2$. What follows is that $a+ib \in \{2,2 \angle \frac{2\pi}{3}, 2 \angle \frac{4\pi}{3}\}$.

*In the equation $a+ib=2$ (which you mistakenly derived), you substituted $ib$ with what you (mistakenly due to the logical fallacy and (1)) thought $b$ was equal to. That is, you forgot to multiply $\sqrt{4-a^2}$ by $i$.

*In the final step you seem to suggest that $\sqrt{x^2-y^2}=x-y$, but this doesn't hold in general.


Finally, the tag linear-algebra doesn't apply to your question.
