<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] $$ (x+y)^3 + (x+y)(x^2-xy+y^2)+33xy=2662 $$
$$ S^3 + S(x^2+y^2-xy) + 33xy=2662 $$
$$ S^3 + S(S^2 - 3xy) + 33xy=2662 $$
$$ 2S^3 + 3xy(11 - S) = 2662 $$
$$ \frac{3}{2}xy(11-S)=11^3 - S^3$$
if $S \ne 11$:
$$ \frac{3}{2}xy = 121 + 11S + S^2 $$
Now try to write down roots of this equation on $S$, add condition of their existence ($D \ge 0$, cause roots are real) and remember that $S = x + y$. Hope this will lead to contradiction.
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