If $a$ and $b$ are the roots of the equation $x^2-px+q=0$, then find the quadratic equation the roots of which are $(a^2-b^2)(a^3-b^3)$ and $a^3b^2+a^2b^3$.

My Try So I tried to write the roots in the form of sum and products of roots of the given equation. The only doubt that I have is, I can convert every term to the sum and product of roots form ( in terms of $p$ and $q$), but I have no idea how to do the same for the term $(a^2-b^2)$.

If there is another better way to solve this problem, let me know. Thanks for devoting this your time.


Hints: $ab=q$, $a+b=p$, $$a^3b^2+a^2b^3=(ab)^2(a+b),$$ $$(a^2-b^2)(a^3-b^3)=(a+b)(a-b)^2((a+b)^2-ab),$$ and $$(a-b)^2 = (a+b)^2-4ab.$$

  • $\begingroup$ It's not $(a-b)^2$ it's $(a^2-b^2)$. $\endgroup$ – Tanuj Aug 13 '17 at 18:15
  • $\begingroup$ $(a-b)(a+b)=a^2-b^2$. $\endgroup$ – Math Lover Aug 13 '17 at 18:16
  • $\begingroup$ Yea that's what I thought. Thanks I'll just give it a try and let you know. $\endgroup$ – Tanuj Aug 13 '17 at 18:16
  • $\begingroup$ I got it. Actually, you know what's better? $(a^3-b^3)$ is actually $(a-b)(a^2+b^2+ab)$.This solves the problem straight away!. But thanks for your time and effort anyway! $\endgroup$ – Tanuj Aug 13 '17 at 18:24
  • $\begingroup$ @user38227 $a-b$ is antisymmetric in the roots (changes sign when you switch the roots), but you get one factor from $a^2-b^2$ and another from $a^3-b^3$ and $(a-b)^2$ is symmetric (because$ (-1)^2=1$). $\endgroup$ – Mark Bennet Aug 13 '17 at 18:24

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