Answer
$x\gt-4$ and $x\ne0$.
Work Step by Step
$2x^3\gt-8x^2\\2x^3+8x^2\gt0\\2x^2(x+4)\gt0$
We know that $(x-k)^2\geq0$ for all $k$.
So here for the inequality to be true, $x+4\gt0\\x\gt-4$ and $x^2\ne0\\x\ne0$
So the solution is $x\gt-4$ and $x\ne0$.
