Answer
$x\gt1$ or $x\lt-1$
Work Step by Step
$x^4\gt x^2\\x^4-x^2\gt0\\x^2(x^2-1)\gt0$
We know that $(x-k)^2\geq0$ for all $k$.
So here for the inequality to be true, $x^2-1\gt0\\x^2\gt1$ and $x^2\ne0\\x\ne0$
$x^2\gt1$ when $x\gt1$ or $x\lt-1$
So the solution is $x\gt1$ or $x\lt-1$
