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

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Answer 1 · 2016-07-29T06:49:15

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

This inequality factors as:

$x^2(x-1) < 9(x-1)$

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

Case $bb(x > 1)$

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

$x^2 < 9$

Hence $-3 < x < 3$

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

Case $bb(x < 1)$

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

$x^2 > 9$

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

So this case gives solutions $x < -3$

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