Solve : $x+y=2010$, $(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942$

Question

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?

Answer 1

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 $$\sqrt[3]{x}+\sqrt[3]{y}=\color{red}{12}$$

Answer 2

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.