# How to solve this inequality with another way?

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?

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Looks fine to me. Why do you want another proof? Is there something in your proof that makes you doubt? –  vonbrand Mar 10 '13 at 17:19

$$\sqrt{(x+2)^3}= \big({\sqrt{x+2}}\big)^3$$
you're left with $t = \sqrt{x+2}$ as the only useful substitution.