Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$. 
Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.

I tried cubing the both sides but things then go very ugly. Are there simpler way to solve it? Thanks.
p.s. The answers are $\pm 6\sqrt 3$.

Set $x+10=t^3$ and $x-10 = u^3$. So we want to solve $t - u = 2$. 
However $t^3 - u^3 = (t-u)^3 + 3tu(t-u) = 20$.
That is $2^3 + 6tu = 8 + 6tu =20$. 
Now we have $tu = 2$. This means $x^2 -100 = 8$ and thus $x = \pm6\sqrt{3}.$
