<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] Rewrite our equation in the following form.
$$(x^3+y^3-11^3+3xy)+(x+y)^3-11^3=0$$
Now, we can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$
Thus, we obtain:
$$(x+y-11)(x^2+y^2+11^2-xy+11x+11y+(x+y)^2+11(x+y)+11^2)=0$$ or
$$(x+y-11)(2x^2+2y^2+xy+22x+22y+242)=0$$ 
and since
$$2x^2+2y^2+xy+22x+22y+242=(x+11)^2+(y+11)^2+x^2+xy+y^2>0,$$
we obtain $$x+y=11$$
Done!
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