<s>[INST] If $x^3+y^3+(x+y)^3+33 xy=2662$, $x,y\in \Bbb R$, find $S=x+y$. 
If $x^3+y^3+(x+y)^3+33 xy=2662$ and $\{x,y\}\subset \Bbb R$, find $S=x+y$.

This question from an olympiad contest. Answer stated: $S=x+y=11$
Tried to develop $(x+y)^3$ to find something useful for the situation, but without success.
 [/INST] Let $x+y=S$ 
$$S^3+(S-x)^3+33 x (S-x)+x^3$$ 
has a derivation 
$$ \frac{d}{dx}(x^3+(S-x)^3+S^3+33 x (S-x) )= -3 (-11 + S) (S - 2 x) $$
So, sollution $ \frac{d}{dx}=0$ leads to $S=11$ or $ S=2x$, i.e. $x=y$
For $x=y$,  $x^3+33 x y+(x+y)^3+y^3=10x^3+33x^2=2662$ has a unique real solution $x(=y)=11/2$ 
For $y=c-x$, function is a ploynomial $2 S^3 + (33 S - 3 S^2) x + (-33 + 3 S) x^2$ with determinant 
$$ -351384 + 31944 S + 1089 S^2 + 66 S^3 - 15 S^4 $$
which is negative except $S=11$ 
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