If #alpha,beta# are the roots of equation #x^2-px+q=0# then find the quadratic equation the roots of which are #(alpha^2-beta^2)(alpha^3-beta^3) & alpha^3 beta^2+ alpha^2 beta^3#?

Given:

#x^2-px+q = (x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta#

We have:

#{ (alpha+beta = p), (alphabeta = q) :}#

Hence:

#(alpha^2-beta^2)(alpha^3-beta^3) = (alpha-beta)^2(alpha+beta)(alpha^2+alphabeta+beta^2)#

#color(white)((alpha^2-beta^2)(alpha^3-beta^3)) = (alpha^2-2alphabeta+beta^2)(alpha+beta)(alpha^2+alphabeta+beta^2)#

#color(white)((alpha^2-beta^2)(alpha^3-beta^3)) = ((alpha+beta)^2-4alphabeta)(alpha+beta)((alpha+beta)^2-alphabeta)#

#color(white)((alpha^2-beta^2)(alpha^3-beta^3)) = (p^2-4q)p(p^2-q)#

#color(white)((alpha^2-beta^2)(alpha^3-beta^3)) = p^5-5p^3q+4pq^2#

and:

#alpha^3beta^2+alpha^2beta^3 = (alpha+beta)(alphabeta)^2 = pq^2#

So the monic quadratic equation with the these two roots can be written:

#0 = x^2-((p^5-5p^3q+4pq^2) + pq^2)x + (p^5-5p^3q+4pq^2)pq^2#

#color(white)(0) = x^2-(p^5-5p^3q+5pq^2)x + (p^6q^2-5p^4q^3+4p^2q^4)#

As #alpha# and #beta# are roots of #x^2-px+q=0#

#alpha+beta=p# and #alphabeta=q#

Therefore #alpha^2+beta^2=p^2-2q#

#alpha^3+beta^3=(alpha+beta)(alpha^2-alphabeta+beta^2)#

= #p(p^2-2q-q)=p(p^2-3q)=p^3-3pq#

and #(alpha^2+beta^2)(alpha^3+beta^3)=alpha^5+beta^5+alpha^2beta^2(alpha+beta)#

or #(p^2-2q)(p^3-3pq)=alpha^5+beta^5+pq^2#

or #alpha^5+beta^5=(p^2-2q)(p^3-3pq)-pq^2#

The quadratic equation whose roots are given is

is #x^2-("sum of roots")x+("product of roots")=0#

Sum of roots is #(alpha^2-beta^2)(alpha^3-beta^3)+alpha^3beta^2+alpha^2beta^3#

= #alpha^5+beta^5-alpha^2beta^3-beta^2alpha^3+alpha^3beta^2+alpha^2beta^3=alpha^5+beta^5#

= #(p^2-2q)(p^3-3pq)-pq^2=p^5-5p^3q+5pq^2#

and product of roots is #(alpha^5+beta^5-alpha^2beta^3-beta^2alpha^3)(alpha^3beta^2+alpha^2beta^3)#

= #(p^5-5p^3q+5pq^2-pq^2)(pq^2)#

= #p^6q^2-5p^4q^3+4p^2q^4#

Hence equation is

#x^2-(p^5-5p^3q+5pq^2)x+(p^6q^2-5p^4q^3+4p^2q^4)=0#