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

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