How can I understand solving the equation?

Question

$$\begin{align} &\left[(\sqrt[4]{p}-\sqrt[4]{q})^{-2} + (\sqrt[4]{p}+\sqrt[4]{q})^{-2}\right] : \frac{\sqrt{p} + \sqrt{q}}{p-q} \\ &= \left(\frac{1}{(\sqrt[4]{p}-\sqrt[4]{q})^{2}}+\frac{1}{(\sqrt[4]{p}+\sqrt[4]{q})^{2}}\right) \times \frac{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})}{\sqrt{p} + \sqrt{q}} \\ &= \frac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}} \times \frac{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})}{(\sqrt{p} + \sqrt{q})}\end{align}$$

How can I get this expression? $$\frac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}}$$ Only this solving I can't understand.

Answer 1

If it helps, try expressing it in another way, recall that we can also express $\sqrt[n]{{a}^m}$ as $a^{m/n}$, so we have

$$\begin{align}\left(\frac{1}{(\sqrt[4]{p}-\sqrt[4]{q})^{2}}+\frac{1}{(\sqrt[4]{p}+\sqrt[4]{q})^{2}}\right)&=\left(\frac{1}{(p^{1/4}-q^{1/4})^2}+\frac{1}{(p^{1/4}+q^{1/4})^2}\right)\\ \end{align} $$

Then, use the properties of quotients $\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}$ and exponents $(a^m)^n=a^{mn}$ to see that

$$\begin{align}\left(\frac{1}{(p^{1/4}-q^{1/4})^2}+\frac{1}{(p^{1/4}+q^{1/4})^2}\right)&=\frac{(p^{1/4}+q^{1/4})^2+(p^{1/4}-q^{1/4})^2}{(p^{1/4}+q^{1/4})^2*(p^{1/4}-q^{1/4})^2}\\ &=\frac{(p^{1/4}+q^{1/4})^2+(p^{1/4}-q^{1/4})^2}{[(p^{1/4}+q^{1/4})*(p^{1/4}-q^{1/4})]^2}\\ &=\frac{(p^{1/4}+q^{1/4})^2+(p^{1/4}-q^{1/4})^2}{(p^{1/2}-q^{1/2})^2}\\ &=\frac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}} \end{align} $$

Answer 2

$$(A+B)^2+(A-B)^2=(A^2+2AB+B^2)+(A^2-2AB+B^2)=2(A^2+B^2)$$ By using this identity, You can simplify the numerator. $$\dfrac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}}=\dfrac{2(\sqrt{p}+\sqrt{q})}{(\sqrt{p} - \sqrt{q})^{2}}$$ Rest will be obvious. Good luck.