<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] There are no solutions in the real numbers.
Plotting $a^2+b^2=4$ on a graph, we get a circle radius 2. Plotting $ab=4$ on the same graph, we get a hyperbola which goes through $(2,2)$ (with the negative side going through $(-2,-2)$) which does not intersect the circle.


A tag of "recreational mathematics" wouldn't usually cause me to go looking for complex solutions unless they were apparent from the setup. However Lythia's answer correctly identifies that the options in the complex plane are restricted to $a$ and $b$ choosing from the $6^{th}$ roots of $-64$. Since $ab=4$, $a$ and $b$ must be complex conjugates. The $a^2+b^2=4$ requirement additionally means that $a$ and $b$ cannot be $\pm 2i$, giving the four answer pairs of $a=\pm \sqrt 3\pm i$ and $b=a^*$.
Allowing complex $a$ and $b$ rescues the question's original analysis and reinstates the answer of $a^3+b^3 = 0$. 
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