6. Find all real numbers $x$ for which
$$
\sqrt{\frac{x^{3}-8}{x}}>x-2
$$
Solution. Firstly, we have to figure out for which real $x$ the left-hand-side makes sense. This happens in two cases:
$$
\left\{\begin{array} { r l } 
{ x ^ { 3 } - 8 } & { \geq 0 } \\
{ x } & { > 0 }
\end{array} \text { and } \left\{\begin{array}{rl}
x^{3}-8 & \leq 0 \\
x & 0$ or $x>2$, then both sides of inequality are positive and it will be equivalent to the inequality
$$
\frac{x^{3}-8}{x}>(x-2)^{2} \text {. }
$$
which simplifies to $-4 x^{2}+4 x+80$. The solution of the latter inequality is $(-\infty,-1) \cup(2, \infty)$. Given that we are under constraint $x>2$, the solution in this case will be $(2, \infty)$.
Answer: $(-\infty, 0) \cup(2, \infty)$.