Answer
The inequality is valid for values between -3 and 0, and values more than 1 (not including them) i.e. $(-3,0)\cap (1,\infty)$
Work Step by Step
First, we are going to find the x-intercepts by equating to zero:
$x^3+2x^2-3x=0$
$x(x^2+2x-3)=0$
$x(x-1)(x+3)=0$
$x_1=-3$
$x_2=0$
$x_3=1$
These are the critical points. We are going to take four values: one less than -3; one between -3 and 0; one between 0 and 1; and one more than 1 to test in the original equation and check if the inequality is true or not:
First test with a value less than -3:
$(-4)^3+2(-4)^2-3(-4)\gt0$
$-64+32+12>0$
$-20>0 \rightarrow \text{ FALSE}$
Second test with a value between -3 and 0:
$(-0.5)^3+2(-0.5)^2-3(-0.5)\gt0$
$-0.125+0.5+1.5>0$
$1.875> \rightarrow \text{ TRUE}$
Third test with a value between 0 and 1:
$(0.5)^3+2(0.5)^2-3(0.5)\gt0$
$0.125+0.5-1.5>0$
$-0.875>0 \rightarrow \text{ FALSE}$
Fourth test with a value greater than 1:
$4^3+2(4)^2-3(4)\gt0$
$64+32-12>0$
$84>0 \rightarrow \text{ TRUE}$
These tests show that the inequality $x^3+2x^2-3x\gt0$ is valid for values between -3 and 0, and values greater than 1 (not including them) i.e. $(-3,0)\cap (1,\infty)$.
