Polynomials hard easy looking problem

Question

If p,q,r are the roots of x^3-6x^2+3x+1=0 , determine the possible values of p^2q+q^2r+r^2p

Answer 1

This is a math competition problem requiring the use of Vieta formulas and some expressions. Since this is a competition problem, I will not put the details of the algebra involved in opening the brackets.

Lets define S=p^2 q + q^2 r + r^2 p and the symmetric version of it T= q^2 p + r^2 q +p^2 r.

From Vieta's formulas we know that:

(1) p+q+r=-(-6)=6

(2) pq+qr+rq=3

(3) pqr=-1

Now lets consider the product of the first two expressions:

18=(p+q+r)*(pq+qr+rp)

After opening the brackets and using pqr=-1 we find that

S+T=21....................................................................................................................................................(i)

OK so I know the sum of S and T. If I can get another expression for S and T, I will have a system of two unknowns and two equations and I can solve for them! So lets consider their product

S*T=p^4 qr + pq^4 r + pqr^4 + p^3 q^3 + q^3 r^3 + r^3 p^3 +3p^2 q^2 r^2

= pqr(p^3+q^3+r^3) + (p^3 q^3 + q^3 r^3 + r^3 p^3 ) + 3 (pqr)^2 ...................................................(ii)

We know pqr=-1 so we need p^3+q^3+r^3 and p^3 q^3 + q^3 r^3 + r^3 p^3

Finding p^3+q^3+r^3 is easy if we consider cubing (1):

216=(p+q+r)^3=p^3+q^3+r^3 +3(S+T)+6pqr=p^3+q^3+r^3+3(21)+6(-1). Note that here I used the expressions from (3) and (i) Thus

p^3+q^3+r^3=159....................................................................................................................................(iii)

Inspired by how we found p^3+q^3+r^3, to find p^3 q^3 + q^3 r^3 + r^3 p^3 we should cube (2):

27=(pq+qr+rp)^3

=p^3 q^3 + q^3 r^3 + r^3 p^3 + 3(p^3 q^2 r + p^3 qr^2 +p^2 q^3 r + p^2 qr^3 +pq^3 r^2 + pq^2 r^3) +6p^2 q^2 r^2

= p^3 q^3 + q^3 r^3 + r^3 p^3 + 3pqr(S+T)+6(pqr)^2

Using that S+T=21 and pqr=-1 (i,3) we have that

p^3 q^3 + q^3 r^3 + r^3 p^3 = 84 ............................................................................................................(iv)

Thus using the expressions from (iii) and (iv) in (ii) we get that

ST=-72 .....................................................................................................................................................(v)

So we have found (i, v) that

S+T=21 and ST=-72.

There are many ways to solve this but it all comes down to the fact that S, T are solutions to

x^2-21x-72=0 which has solutions 24 and -3.

Note that the desired expression S can be either of these values since the difference between S and T is simply the order in which you take the roots.