<s>[INST] Skanavi 2.003, Difference in answers, which is right? In Skanavi book i have a exercise to simplify an equation
$$\begin{align}
((\sqrt[4]{p} - \sqrt[4]{q})^{-2} + (\sqrt[4]{p} + \sqrt[4]{q})^{-2}) : \frac{\sqrt p + \sqrt q}{p-q}
\end{align}$$
Solving:
Replacing:
$$ a =\sqrt[4]{p} $$
 $$ b=\sqrt[4]{q} $$
Equation is changed:
$$
((a - b)^{-2} + (a + b)^{-2}): \frac{a^2+b^2}{a^4-b^4}
$$
$$ A = ((a - b)^{-2} + (a + b)^{-2})$$ 
$$ B = \frac{a^2+b^2}{a^4-b^4}$$ 
$$ B =  \frac{a^2+b^2}{a^4-b^4} => B = \frac{a^2+b^2}{(a^2-b^2)(a^2+b^2)} =>  B = \frac{1}{a^2 - b^2}$$
$$ A = ((a - b)^{-2} + (a + b)^{-2}) => A = \frac{1}{(a - b)^{2}} + \frac{1}{(a + b)^{2}} => A = \frac{(a + b)^2 + (a-b)^2}{(a^2 - 2ab + b^2)(a^2 + 2ab + b^2)} => A = \frac{a^2 + 2ab + b^2 + a^2 - 2ab + b^2}{a^4 + 2a^2b^2 + b^4 - 2a^2b - 4a^2b^2 - 2ab^2 + a^2b^2+2ab^2+b^4} => A =\frac{2a^2 + 2b^2}{a^4 - 2a^2b^2 + b^4} => A = \frac{2(a^2+b^2)}{(a^2-b^2)^2}$$
$$ A : B $$
$$ \frac{2(a^2+b^2)}{(a^2-b^2)^2} : \frac{1}{a^2 - b^2}  =  \frac{2(a^2+b^2)(a^2 - b^2)}{(a^2-b^2)^2} = \frac{2(a^2+b^2)}{(a^2-b^2)}$$
My Answer is:
$$
\frac{2(\sqrt p +\sqrt q)}{(\sqrt p-\sqrt q)}
$$
Books answer is:
$$
\frac{2(\sqrt p +\sqrt q)^2}{(p- q)}
$$
Where is my mistake?
 [/INST] Since you have fourth roots you can do the following, to eliminate roots from the denominator.
$$\frac {2(a^2+b^2)}{a^2-b^2}=\frac {2(a^2+b^2)}{a^2-b^2}\cdot\frac {a^2+b^2}{a^2+b^2}=\frac {2(a^2+b^2)^2}{a^4-b^4}$$
Substitute your values into this and you will see that the two answers are equal to each other.

Alternatively, without any short cuts $$\frac{\sqrt p+\sqrt q}{\sqrt p-\sqrt q}-\frac {(\sqrt p+\sqrt q)^2}{p-q} =$$$$\frac {p\sqrt p+p\sqrt q-q\sqrt p-q\sqrt q-p\sqrt p-2p\sqrt q-q\sqrt p+p\sqrt q+2q\sqrt p+q\sqrt q}{(\sqrt p-\sqrt q)(p-q)}=0$$
</s>