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.

To give a different perspective, I can't find anywhere in the problem statement which states $a$ and $b$ must be real. Therefore, assume $a,b \in \mathbb{C}$. Then the equation $a^3+b^3=0$ along with the condition $ab=4$ implies, by multiplying both sides by either $a^3$ or $b^3$ that $a^6+64 = 0$ and $b^6+64=0$. Therefore, 
$$ a = \pm\sqrt[6]{-64} = \pm2\sqrt[6]{-1},\\  b= \pm\sqrt[6]{-64} = \pm2\sqrt[6]{-1}.$$
As $a$ and $b$ are multiples of sixth roots of unity, there are six choices for each. Not all choices are valid since the condition $a^2+b^2=4$ must also be met. For example 
$$a=b=\sqrt{3}+i$$
does not work since then $a^2+b^2=4+4\sqrt{3}i$. However, the choice 
$$a=\sqrt{3}+i, b=\sqrt{3}-i$$
works fine.
