We must have: $x\neq 0$ $\frac{x^3 - 8}{x} \geq0$ $x-2\geq 0$ Therefore we must have $x\geq 2$ If $x-2\geq1$ then we can solve for $\frac{x^3-8}{x}>(x-2)^2$ which yields $x>2$ If $0\leq x-2<1$ then we can solve for $\frac{x^3-8}{x}<(x-2)^2$ which yields $-1<x<2$ which are outside the domain of the inequality Solution: $(2, \infty)$

Lankou wrote: We must have: $x\neq 0$ $\frac{x^3 - 8}{x} \geq0$ $x-2\geq 0$ Therefore we must have $x\geq 2$ If $x-2\geq1$ then we can solve for $\frac{x^3-8}{x}>(x-2)^2$ which yields $x>2$ If $0\leq x-2<1$ then we can solve for $\frac{x^3-8}{x}<(x-2)^2$ which yields $-1<x<2$ which are outside the domain of the inequality Solution: $(2, \infty)$ Not quite. The step from $\frac{x^3-8}{x} \ge 0$ to $x\ge 2$ is not valid because you can't clear the denominator if $x$ is negative. From $\frac{x^3-8}{x} \ge 0$, you get $x \ge 2$ or $x<0$. If $x<0$, the left side is positive and the right side is negative, so all of those values work. If $x\ge 2$, we need to verify that $\frac{x^3-8}{x} > (x-2)^2$, which holds for all $x>2$. So, the answer is $(-\infty, 0) \cup (2,\infty)$.

Thank you for the correction