<s>[INST] 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 $\pm 6\sqrt 3$.
 [/INST] $\displaystyle\sqrt[3]{x+10}-\sqrt[3]{x-10}=2---(1)$
Let $y=x+10$
$(1)$ becomes
$\displaystyle\sqrt[3]{y}-\sqrt[3]{y-20}=2$
$\displaystyle\sqrt[3]{y}=2+\sqrt[3]{y-20}$
Cube both sides to get
$\displaystyle{y}=8+12  (y-20)^{\frac{1}{3}}+6(y-20)^{\frac{2}{3}}+y-20$
$\displaystyle{0}=-12+ 6  (y-20)^{\frac{1}{3}}(2+(y-20)^{\frac{1}{3}})$
$\displaystyle{12}= 6  (y-20)^{\frac{1}{3}}(y)^{\frac{1}{3}}$
$\displaystyle{2}= (y-20)^{\frac{1}{3}}(y)^{\frac{1}{3}}$
Cube both sides
$\displaystyle{8}= (y-20)(y)$
$\displaystyle{y^2-20y-8}= 0$
Solving the above equation yields $y=10\pm 6\sqrt{3}$ which implies $x=\pm 6\sqrt{3}$.
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