Simplify $\sqrt{\frac{\sqrt[4]{x^3}-8}{\sqrt[4]{x}-2}+2\sqrt[4]{x}}\left(\frac{\sqrt[4]{x^3}+8}{\sqrt[4]{x}+2}-\sqrt{x}\right)$ Simplify $$\sqrt{\dfrac{\sqrt[4]{x^3}-8}{\sqrt[4]{x}-2}+2\sqrt[4]{x}}\left(\dfrac{\sqrt[4]{x^3}+8}{\sqrt[4]{x}+2}-\sqrt{x}\right)$$
Is it a good idea to simplify the square root with the common denominator $\sqrt[4]{x}-2$. I tried it and it seemed useless at the end. Any hints would be appreciated. Thank you!

Start with setting $\sqrt[4]{x} = u$ (also mentioned by $\textit{Achilles hui}$  in the comments).

We get
$$L= \sqrt{\frac{u^3-8}{u-2}+2u}\left(\frac{u^3+8}{u+2}-u^2\right)$$
Using the identities for sum and differences of cubes we have
$$L=\sqrt{(u^2+2u+4)+2u}((u^2-2u+4)-u^2)$$
$$L=(u+2)\cdot ( -2(u-2))=-2(u^2-4)=-2(\sqrt{x}-4)$$
