## Precalculus (10th Edition)

$(-\infty,-1) \cup (0,1)$
The equation according to the text: $x^3\lt x\\ x^3-x\lt0\\(x-1)x(x+1)\lt0$ The zeros: $x=0$ or $x+1=0\\x=-1\\$ or $x-1=0\\x=1$. Then, according to the table (where I used the zeros ($x=-1,0,1$) to divide the number line into different parts), the solution set: $(-\infty,-1) \cup (0,1)$ because the function is negative in this interval.