How to solve the following equation $\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$

Question

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.

Answer 1

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, $x\in\left\{-\frac{1}{3},0\right\}$.

Answer 2

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 $x=-\frac{1}{3}$.