Prove that $\frac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\frac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}$ Prove that $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\dfrac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}, a>0, x>0,x\ne a.$$ My try $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\dfrac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=\dfrac{\sqrt{a}^2-\sqrt{x}^2}{\sqrt{a}-\sqrt{x}}-\dfrac{a+\sqrt[4]{ax^3}-\sqrt[4]{a^3x}-\sqrt[4]{a^2x^2}}{\sqrt{a}+\sqrt[4]{ax}}$$ Can we do something more directly to arrive at the solution? Thank you!
 A: $$
\begin{aligned}
\frac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\frac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right) &= \frac{\left(\sqrt{a}\right)^2-\left(\sqrt{x}\right)^2}{\sqrt{a}-\sqrt{x}}-\frac{\left(\sqrt[4]{a}\right)^4+\sqrt[4]{a}\left(\sqrt[4]{x}\right)^3}{\left(\sqrt[4]{a}\right)^2+\sqrt[4]{a}\sqrt[4]{x}}+\sqrt[4]{ax} = \\
&=\frac{\left(\sqrt{a}-\sqrt{x}\right)\left(\sqrt{a}+\sqrt{x}\right)}{\sqrt{a}-\sqrt{x}}-\frac{\sqrt[4]{a}\left((\sqrt[4]{a})^3+(\sqrt[4]{x})^3\right)}{\sqrt[4]{a}\left(\sqrt[4]{a}+\sqrt[4]{x}\right)}+\sqrt[4]{ax} = \\
&= \sqrt{a}+\sqrt{x}-\frac{(\sqrt[4]{a}+\sqrt[4]{x})((\sqrt[4]{a})^2-\sqrt[4]{a}\sqrt[4]{x}+(\sqrt[4]{x})^2)}{\sqrt[4]{a}+\sqrt[4]{x}}+\sqrt[4]{ax} = \\
&= \sqrt{a}+\sqrt{x}-(\sqrt{a}-\sqrt[4]{ax}+\sqrt{x})+\sqrt[4]{ax} = \\
&= \sqrt{a}+\sqrt{x}-\sqrt{a}+\sqrt[4]{ax}-\sqrt{x}+\sqrt[4]{ax} = 2\sqrt[4]{x} 
\end{aligned}
$$
A: Just to make things easier to type let $c =\sqrt[4]a$ and $y=\sqrt[4]x$ so we are asked to prove
$\frac {c^4 - y^4}{c^2-y^2} -(\frac {c^4 + cy^3}{c^2 + cy} - cy)= 2cy$.
so lets do that
$\frac {c^4 - y^4}{c^2-y^2} -(\frac {c^4 + cy^3}{c^2 + cy} - cy)=$
$\frac {(c^2-y^2)(c^2+y^2)}{c^2-y^2} -\frac {c(c^3+ y^3)}{c(c+y)} + cy=$
$c^2+cy+y^2 -\frac {c^3+y^3}{c+y}=$
$c^2 + cy + y^2 - \frac {(c+y)(c^2 -cy+y^2)}{c+y}=$
$c^2 + cy + y^2 -(c^2 -cy +y^2) = 2cy$
That's it.
