Help solving $\sqrt[3]{x^3+3x^2-6x-8}>x+1$

Question

I'm working through a problem set of inequalities where we've been moving all terms to one side and factoring, so you end up with a product of factors compared to zero. Then by creating a sign diagram we've determined which intervals satisfy the inequality.

This one, however, has me stumped: $$\sqrt[3]{x^3+3x^2-6x-8}>x+1$$ I managed to do some factoring, and think I might be on the right path, though I'm not totally sure. Here's what I have so far, $$\sqrt[3]{x^3+3x^2-6x-8}-(x+1)>0\\ \sqrt[3]{(x-2)(x+4)(x+1)}-(x+1)>0\\ \sqrt[3]{(x-2)(x+4)}\sqrt[3]{(x+1)}-(x+1)>0\\ \sqrt[3]{(x-2)(x+4)}(x+1)^{1/3}-(x+1)>0\\ (x+1)^{1/3}\left(\sqrt[3]{(x-2)(x+4)}-(x+1)^{2/3}\right)>0$$ Is there a way to factor it further (the stuff in the big parentheses), or is there another approach I'm missing?

Update: As you all pointed out below, this was the wrong approach.

This is a lot easier than you are making it.

How very true :) I wasn't sure how cubing both sides would affect the inequality. Now things are a good deal clearer. Thanks for all your help!

Answer 1

Note that the cubic function is strictly increasing, so $z>y$ if and only if $z^3>y^3$.

Cube both sides of the inequality you want to solve, and you can reduce it to a linear inequality.

Answer 2

Hint:$$\sqrt[3]{x^3+3x^2-6x-8}>x+1$$$${x^3+3x^2-6x-8}>(x+1)^3$$

Answer 3

This is a lot easier than you are making it. Simply cube both sides (cubes preserve sign, so this introduces no complications):

$$x^3+3 x^2-6 x-6 > x^3+3 x^2+3 x+1$$

The cube and square factors cancel, and we simply get that $x < -1$.

Answer 4

Hint: you may want to remove the root first, by noticing that $f(y)=y^3$ is a monotonic increasing function.