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.
 A: 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!
A: $$ (x+y)^3 + (x+y)(x^2-xy+y^2)+33xy=2662 $$
$$ S^3 + S(x^2+y^2-xy) + 33xy=2662 $$
$$ S^3 + S(S^2 - 3xy) + 33xy=2662 $$
$$ 2S^3 + 3xy(11 - S) = 2662 $$
$$ \frac{3}{2}xy(11-S)=11^3 - S^3$$
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.
A: 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 $S=11$ 
