Find all $x \in \mathbb{R}$ such that:
$\sqrt[3]{x^6+7x^3}+(x-1)(x+1)=\sqrt{x^4+8x}-1$
My try: Expanding, it is:
$\sqrt[3]{x^6+7x^3}+x^2=\sqrt{x^4+8x}$
By inspection, $x=0$ is a solution. If I divide by $x$:
$\sqrt[3]{x^3+7}+x=\sqrt{x^2+\frac{8}{x}}$
$x = 1$ is solution. I don't know how to find others.
Firstly, the square root is defined only if $x\leq -2$ or $x\geq 0$. Clearly, $x=0$ is a solution. For $x\neq 0$, the equation is equivalent with:
$$\sqrt[3]{x^6+7x^3}=\sqrt{x^4+8x}-x^2=\frac{8x}{\sqrt{x^4+8x}+x^2}$$
or
$$\sqrt[3]{x^3+7}\left(\sqrt{x^4+8x}+x^2\right)=8$$
If $x\geq 0$, the left hand side is an increasing function, so we can have at most one solution, which is $x=1$. If $x\leq -2$, the left hand side is negative, so no solution.
In conclusion, the only solutions are $\{0,1\}$.