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?

From what you've done, we have that
$$(1)\iff 2010=(u+v)^3-3uv(u+v)\tag3$$
and that
$$(2)\iff uv(u+v)=2942-11(u+v)^2-121(u+v)\tag4$$
From $(3)(4)$, we have
$$2010=(u+v)^3-3(2942-11(u+v)^2-121(u+v)),$$
i.e.
$$t^3+33t^2+363t-10836=0\tag5$$
where $t=u+v$.
Now checking if a divisor of $10836=2^2\times 3^2\times 7\times 43$ is a solution, we see that $t=12$ is a solution to have
$$(5)\iff (t-12)(t^2+45t+903)=0$$
where $t^2+45t+903\gt 0$ since $45^2-4\cdot 1\cdot 903\lt 60^2-4\times 900=0$.
It follows from this that $$\sqrt[3]{x}+\sqrt[3]{y}=\color{red}{12}$$
