# Is there a more systematic approach to solving this inequality?

$$x^3>x$$

Steps I took:

$$x^{ 3 }-x>0$$

$$x(x^{ 2 }-1)>0$$

$$x(x-1)(x+1)>0$$

Now I see that all three linear factors must equal a positive value when multiplied.

I took each linear factor and set each to either greater than or less than $0$ since the solution set can either be all positive or two negative and one positive.

$$x>0\quad or\quad x< 0,\quad x>1\quad or\quad x<1,\quad x>-1\quad or\quad x <-1$$

Now where do I go from here? Or am I doing this all wrong?

By solving corresponding equality, we get 3 values -1,0,1. This will give us 4 intervals $(-\infty,-1), (-1,0), (0,1), (1, \infty)$. By checking on this intervals, the inequality is satisfied for intervals $(1,\infty)$ and $(-1,0)$
Either all $3$ factors, $x+1$, $x$, and $x-1$ are positive or one and only one is positive.
Either $x>1$, in which case all $3$ factors are positive or $-1<x<0$, in which case only one, $x+1$, is positive and the other two are negative. In both cases the inequality holds.