<s>[INST] Using the factorization of a quadratic equation If $a,b$ are the roots of the equation $x^2 -px +q=0$ and $a_1, b_1$ are the roots of the equation $x^2-qx +p=0$ then find the equation of quadratic whose roots are $\frac{1}{a_1b} + \frac{1}{ab_1}$ and $\frac{1}{aa_1} + \frac{1}{bb_1}$.
I know that the equation $Qx^2 + bx +c =0$ whose roots are $e,d$ can be expressed as $Q(x-e)(x-d)=0$.  What I'm confused about in this question stated above is that when I try to write the new asked quadratic equation what do I take the $Q$ term to be?
 [/INST] $$\left(\frac{1}{a_1b} + \frac{1}{ab_1}\right)+\left(\frac{1}{aa_1} + \frac{1}{bb_1}\right)=\frac{(a_1+b_1)(a+b)}{a_1b_1ab}=\frac{pq}{pq}=1.$$
$$\left(\frac{1}{a_1b} + \frac{1}{ab_1}\right)\left(\frac{1}{aa_1} + \frac{1}{bb_1}\right)=\frac{(ab_1+a_1b)(aa_1+bb_1)}{p^2q^2}=$$
$$=\frac{a^2q+b_1^2p+a_1^2p+b^2q}{p^2q^2}=\frac{(q^2-2p)q+(p^2-2q)p}{p^2q^2},$$
which by the Viete's theorem gives the answer:
$$x^2-x+\frac{p^3+q^3}{p^2q^2}-\frac{4}{pq}=0.$$
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