<s>[INST] Find all possible values of $ a^3 + b^3$ if $a^2+b^2=ab=4$. Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$. 
If $a=b=0$, it is conflict with $a^2+b^2=ab=4$.
If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one negative. This contradict with $ab$=4.
I don't know the solution. 
Any help please.
 [/INST] It is given that $a^{2} + b^{2} = ab =4$ and the goal is to determine all values of $a^{3} + b^{3}$. Now, it is seen that $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2}) = 0$. So far the set of equations becomes
\begin{align}
a^{3} + b^{3} &= 0 \\
a^{2} + b^{2} &= 4 \\
ab &= 4.
\end{align}
From the first, multiply by $a^{3}$ to obtain $0 = a^{6} + (ab)^{3} = a^{6} + 2^{6}$ which yields $a = \pm 2 e^{\pi i/6 \pm 2n \pi i}$. Since $ab=4$ then yields $b = \pm 2 e^{-\pi i/6 \pm 2n\pi i}$. In both cases $n \geq 0$. The possible values of $a$ are
\begin{align}
a \in \{ 2 e^{\pi i/6 + 2n \pi i}, 2 e^{\pi i/6 - 2n \pi i}, - 2 e^{\pi i/6 + 2n \pi i}, - 2 e^{\pi i/6 - 2n \pi i} \}
\end{align}
and for $b$
\begin{align}
b \in \{ 2 e^{-\pi i/6 + 2n \pi i}, 2 e^{-\pi i/6 - 2n \pi i}, - 2 e^{-\pi i/6 + 2n \pi i}, - 2 e^{-\pi i/6 - 2n \pi i} \}
\end{align}
In view of these two sets of values it is seen that the sets must be taken in order of signs, ie the sets $(a, b) \in \{(+,+), (-,-)\}$,
\begin{align}
a^{3} + b^{3} &= 2 (\pm 2)^{3} \cos\left(\frac{\pi}{2}\right) = 0 \\
a^{2} + b^{2} &= 2 (\pm 2)^{2} \cos\left(\frac{\pi}{3}\right) = 4 \\
ab &= (\pm 2)^{2} = 4.
\end{align} 
</s>