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`