Simultaneous equations

p+q+r=2
p^2+q^2+r^2=6
p^3+q^3+r^3=8
(p+q+r)(p^2+q^2+r^2-pq-pr-rq)=p^3+q^3+r^3-3pqr

can anyone give me a clue about the best way to solve these simultaneous to give results for p, q and r. I'm lost so ideas as i cant seem to eliminate a second variable regardless of what i try...

Reply 1

Farhan.Hanif93

18

Original post by Goods

p+q+r=2
p^2+q^2+r^2=6
p^3+q^3+r^3=8
(p+q+r)(p^2+q^2+r^2-pq-pr-rq)=p^3+q^3+r^3-3pqr

can anyone give me a clue about the best way to solve these simultaneous to give results for p, q and r. I'm lost so ideas as i cant seem to eliminate a second variable regardless of what i try...

It's worth pointing out that the 4th equation is actually an identity [but is still useful for this problem - otherwise I presume it wouldn't be there!].

As a hint, observe that the cubic polynomial

x^3 -(p+q+r)x^2 + (pq+pr+rq)x -pqr

has roots

p,q,r

(if this isn't immediately obvious, expand

(x-p)(x-q)(x-r)

). Hence, if you can find these coefficients using what the system of simultaneous equations tells you, you will have a cubic equation that will yield precisely the values of

p,q

and

r

that solve the system -- Use the first two equations to determine

pq+pr+rq

; and then use this and the 'fourth' equation to determine

pqr

.

You'd then want to ask yourself whether you have every solution - can you see a way to confirm that the implication goes both ways here?

Reply 2

brianeverit

9

Original post by Goods

p+q+r=2
p^2+q^2+r^2=6
p^3+q^3+r^3=8
(p+q+r)(p^2+q^2+r^2-pq-pr-rq)=p^3+q^3+r^3-3pqr

can anyone give me a clue about the best way to solve these simultaneous to give results for p, q and r. I'm lost so ideas as i cant seem to eliminate a second variable regardless of what i try...