<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] As has been observed, one approach to this problem is to solve the equations over the complex numbers, and then check directly that $a^3+b^3=0$ in all cases.  However, it's both useful and more general to observe that the conclusion can be reached without actually solving the equations: since $a^2+b^2=a b$, we have

$(*)\ \ a^3+b^3=(a+b)(a^2-a b + b^2)=0.$

The problem makes sense over any field  -- in fact in any commutative ring with $1$ (this condition is only needed so that "$4$" makes sense.)  In addition to being simple, an advantage of the solution $(*)$ is that it works regardless of the setting.  This is relevant, since the problem did not specify a domain for $a$ and $b$, so a solution based on calculating the roots in a particular context involves an inherent assumption.
For example, in the field ${\Bbb F}_{37}$ (i.e., the integers mod $37$), the equations have the four solutions $(a,b)$=$(9,21)$, $(16,28)$, $(21,9)$, and $(28,16)$.  You can check that $a^3+b^3=0$ in each case, but you're back to square one when faced with ${\Bbb F}_{61}$.
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