TITLE: How to solve this inequality with another way?
QUESTION [5 upvotes]: I want to solve this inequality $$x^3-3 x^2+2 \sqrt{(x+2)^3}-6
   x\geqslant 0.$$
I tried. Put $t = \sqrt{x + 2}$. Then, we get
$$t^6+2t^3-9t^4+18t^2-8\geqslant 0.$$ 
Equavalent to 
$$(-2 + t)^2 (1 + t)^2 (-2 + 2 t + t^2)\geqslant 0.$$
From this inequality, we have $x\geqslant 2
   \left(1-\sqrt{3}\right).$
How to solve this inequality with another way?

REPLY [2 votes]: As mentioned in the comment, your approach looks fine. Specifically, to approach a solution you need to "remove" the surd, and noting that:
$$  \sqrt{(x+2)^3}= \big({\sqrt{x+2}}\big)^3 $$
you're left with $ t = \sqrt{x+2} $ as the only useful substitution.
I should add: your "Equavalent to" step is really a big one, but if you showed that working, I see nothing to change.