<s>[INST] How to solve the following equation $\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$ I am trying to solve this equation:
$$\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$$
I would like to get some advice, how to solve it. 
Thanks.
 [/INST] First note that $x=0$ is a solution. Now we consider $x\neq 0$ and we divide by $\sqrt[3]{x}$. We get $\sqrt[3]{1+\frac{3}{x}}+1=\sqrt[3]{8+\frac{3}{x}}$, so lets define $y=\frac{3}{x}$ and write
$$1+\sqrt[3]{1+y} = \sqrt[3]{8+y}.$$
Now we cube both sides to obtain
$$1+3\sqrt[3]{1+y}+3\sqrt[3]{1+y}^2+1+y = 8+y$$
or
$$\sqrt[3]{1+y}+\sqrt[3]{1+y}^2 = 2.$$
We complete the square to obtain
$$\left(\sqrt[3]{1+y}+\frac{1}{2}\right)^2=\frac{9}{4}$$
or
$$\sqrt[3]{1+y} = \frac{-1\pm 3}{2}.$$
Hence $y = 0$ or $y=-9$.
There is no $x$ corresponding to $y=0$, but corresponding to $y=-9$ we have $x=-\frac{1}{3}$.
Concluding, $x\in\left\{-\frac{1}{3},0\right\}$.
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