<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] W'll use $$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2.$$
Since $x+3=x=-3-8x$ is impossible, our equation is equivalent to
$$x+3+x-3-8x+3\sqrt[3]{x(x+3)(8x+3)}=0$$ or
$$\sqrt[3]{x(x+3)(8x+3)}=2x$$ or
$$x((x+3)(8x+3)-8x^2)=0,$$ 
which gives $x=0$ or $x=-\frac{1}{3}$.
</s>