<s>[INST] Solve : $x+y=2010$, $(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942$ 
Given$$x+y=2010,\\(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942,$$
  find the value of $\sqrt[3]{x}+\sqrt[3]{y}$.

My attempts:
Let $u=\sqrt[3]{x} , v=\sqrt[3]{y}$.
$(u+v)(u^2-uv+v^2)=2010 \tag{1}$
$(u+v)(uv+11u+11v+121)=2942
\tag{2}$
$(1)+(2)$:
$$(u+v)(u^2+v^2+11u+11v+121)=4952$$
How to proceed?
 [/INST] u and v look quite similar, so I will just use a and b instead:
$$3(a+b)(ab+11a+11b+121)+(a+b)(a^2-ab+b^2)=3\times2942+2010=10836$$
$$(a+b)(a^2+2ab+b^2+33a+33b+363)=10836$$
$$(a+b)((a+b)^2+33(a+b)+363)=10836$$
$$(a+b)^3+33(a+b)^2+363(a+b)-10836=0$$
Solving for $a+b$, the real root to this is $a+b=12$, so the answer is 12. I'm not sure if this is the desired approach, but it works.
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