Solve the inequality $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$

Question

So am trying to solve this inequality, $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$, the problem is of course for values $0<a<1$, I tried working through it multiple times but the solutions i found, which they are $\sqrt{1-\sqrt{1-a}}<x<1 $ are the wrong ones when I try wolfram alpha, can anyone show me the right way to do it please?

Answer 1

You obviously need $x > 0$ and further have $$\sqrt[3]{(1+\sqrt x)^3} - \sqrt[3]{(1-\sqrt x)^3} > x+a$$

$\iff 1-a > (\sqrt x - 1)^2 \iff \sqrt{1-a} > |\sqrt x - 1|$

For $x \in (0, 1)$, we have this as $ \sqrt{1-a} > 1-\sqrt x \implies x > (1-\sqrt{1-a})^2 $

For $x \geqslant 1$, we have this as $ \sqrt{1-a} > \sqrt x - 1 \implies (1+\sqrt{1-a})^2 > x $

Thus the solution set is $x \in \left((1-\sqrt{1-a})^2, (1 + \sqrt{1-a})^2\right)$.