What is the value of p^3/(q^4+r^4) +q^3/(p^4+r^4) +r^3/(p^4+q^4) from this equation?

Question

The roots of the cubic equation
x^3 -4x^2 +5x -6 =0
are p,q, and r.

What is the value of
p^3/(q^4+r^4) +q^3/(p^4+r^4) +r^3/(p^4+q^4) ?

Answer 1

[p^3(p^4+q^4)(p^4+r^4)+q^3(p^4+q^4)(q^4+r^4)+r^3(r^4+q^4)(p^4+r^4)]

/(p^4+q^4)(p^4+r^4)(q^4+r^4)

bottom is p^8q^4+p^8r^4+q^8p^4+ q^8r^4+r^8q^4+r^8p^4+ 2p^4r^4q^4

= p^4q^4(p^4+q^4+r^4) +p^4r^4(p^4+q^4+r^4) +q^4r^4(p^4+q^4+r^4) -p^4q^4r^4

=(p^4+q^4+r^4) (p^4q^4 + p^4r^4 +q^4r^4) - p^4q^4r^4

numerator is p^7(p^4+q^4+r^4)+ p^3q^4r^4 + q^7(p^4+q^4+r^4) + p^4q^3r^4

+ r^7(p^4+q^4+r^4) +p^4q^4r^3

=(p^7+ q^7+r^7) (p^4+q^4+r^4) + p^3q^3r^3 (pq+qr+pr)



p+q+r=4, pqr =5, pq+qr+pr=6

p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=4

p^3+q^3+r^3=(p+q+r)^3- 3(p+q+r) (pq+pr+qr) +3pqr= 7

p^4+q^4+r^4 = (p^2+q^2+r^2)^2 - 2(p^2q^2+ r^2p^2+ q^2r^2)

= (p^2+q^2+r^2)^2 - 2[ (pq+qr+pr)^2 -2prq(p+q+r)]

= 16-2[-4]=24

(p^2q^2+ r^2p^2+ q^2r^2)=-4

p^7+q^7+r^7 = p^3+q^3+r^3 * p^4+q^4+r^4 -[

p^3q^4+p^4q^3 + p^3r^4+p^4r^3+ r^3q^4+r^4q^3]

= p^3+q^3+r^3 * p^4+q^4+r^4 -

(p^3q^3+p^3r^3+q^3r^3) (p+q+r)+ (p^2q^2+q^2r^2+p^2r^2) (pqr)

=p^3+q^3+r^3 * p^4+q^4+r^4 -

{(pq+qr+pr) (p^2q^2+ r^2p^2+ q^2r^2)- pqr [(pq+qr+pr) (p+q+r)-3pqr]}

(p+q+r)+ (p^2q^2+q^2r^2+p^2r^2) (pqr)

7*24- {6*-4- 5[6*4-3*5]}4+-4*5 =168- {-24-5[9 ]}4-20

=424

p^4q^4 + p^4r^4 +q^4r^4= (p^2q^2 + p^2r^2 +q^2r^2)^2 -2p^2q^2r^2 (p^2+q^2+r^2)

=16-2*5*5*4= -184

(424*24+5^3*6)/(24*-184-5^4)

10926/-5041 = -2/1674

Answer 2

For interest, I’ll do this without extracting the real root. Mike’s method is the quickest (but he made

some slips). This way might be used if you can’t extract easily.



∑ p³/(q⁴+r⁴) = { ∑ p³(q⁴+p⁴)(r⁴+p⁴) } / ((q⁴+r⁴)(r⁴+p⁴)(p⁴+q⁴) }





∑ p³(q⁴+p⁴)(r⁴+p⁴) = ∑ { p³q⁴r⁴ + p⁷(q⁴+r⁴+p⁴) }



Derivation of various sums used is given at the end



∑ p³q⁴r⁴ = (pqr)³∑qr = 6³*5 = 1080



∑ p⁷(p⁴+q⁴+r⁴) = ∑p⁷ * ∑p⁴ = 2174*82



∴ ∑ p³(q⁴+p⁴)(r⁴+p⁴) = 179348





(q⁴+r⁴)(r⁴+p⁴)(p⁴+q⁴) = (82−p⁴)(82−q⁴)(82−r⁴)



= 82³ − 82²*∑p⁴ + 82*∑p⁴q⁴ − (pqr)⁴ = 82*97 − 6⁴ = 6658



∴ ∑ p³/(q⁴+r⁴) = 179348 / 6658 = 89674/3329







………………………………………



p³ = 4p² − 5p + 6 = 0 … (i)



∑p=4, ∑pq=5, pqr=6, ∑p² = (∑p)²−2∑pq = 6



Let S(n) = ∑pⁿ



Then multiply (i) by pⁿ and sum to get for n≥0 : S(n+3) = 4S(n+2)−5S(n+1)+6S(n)



Applying repeatedly using S(1)=5, S(2)=6 gives the results



S(3)=22, S(4)=82, S(5)=254, S(6)=738, S(7)=2174, S(8)=6530



Also 2∑ p⁴q⁴ = (∑p⁴)² − ∑ p⁸ → ∑ p⁴q⁴ = ½(82²−6530) = 97



………………………………………



edit : A more elegant way to do this is to develop the cubic equation (in y) whose roots correspond to the three given values involved in the sum. Here this involves eliminating x from the equations



x³−4x²+5x−6=0 and y=x³/(82−x⁴)



This comes to −3329y³ + 89674y² + 5639y + 108 = 0



More than you requested but does lead to an interesting general approach for problems of this type.



………………………………………

Answer 3

the roots are 3,

.5+isqrt(7)/2.

.5-isqrt(7)/2

Answer 4

p+q+r = 4 -------------- (1)

pq+qr+rp = 5-----------(2) and

pqr = 6