# How do you solve x^3-x^2<9x-9?

Jul 29, 2016

$x < - 3$ or $1 < x < 3$

#### Explanation:

This inequality factors as:

${x}^{2} \left(x - 1\right) < 9 \left(x - 1\right)$

If $x = 1$ then both sides are $0$ and the inequality is false.

Case $\boldsymbol{x > 1}$

$\left(x - 1\right) > 0$ so we can divide both sides by $\left(x - 1\right)$ to get:

${x}^{2} < 9$

Hence $- 3 < x < 3$

So this case gives solutions $1 < x < 3$

Case $\boldsymbol{x < 1}$

$\left(x - 1\right) < 0$ so we can divide both sides by $\left(x - 1\right)$ and reverse the inequality to get:

${x}^{2} > 9$

Hence $x < - 3$ or $x > 3$

So this case gives solutions $x < - 3$