## abtrehearn 4 years ago Suppose P(x) = x^3 - a x^2 + b x - c has three distinct real roots, r1, r2, and r3. What cubic polynonial with 1 as its leading coefficient has r1^4, r2^4, and r3^4 as its roots?

1. jamesm

(x - r1)(x - r2)(x - r3) = x^3 - ax^2 + bx - c x^3 - (r1 + r2 + r3)x^2 + (r1r2 - r3r1 - r3r2)x - r1r2r3 = x^3 - ax^2 + bx - c thus r1 + r2 + r3 = a [1] r1r2 - r3r1 - r3r2 = b [2] r1r2r3 = c [3] (x - r1^4)(x - r2^4)(x - r3^4) = x^3 - (r1^4 + r2^4 + r3^4)x^2 + (r1^4*r2^4 - r3^4*r1^4 - r3^4*r2^4)x - r1^4*r2^4*r3^4 We can see from equations [1], [2] and [3], if we raise both sides of [3] to the power of 4: r1^4 * r2^4 * r3^4 = c^4 so last term of the polynomial is -c^4

2. jamesm

struggling to see how to express the other terms in terms of a, b and c

3. abtrehearn

Looks good so far.

4. myininaya

5. abtrehearn

That's right.

6. abtrehearn

The coefficients of the polynomial we're looking for are to be expressed in terms of a, b, and c.

7. jamesm

requires an algebraic trick that i can not see...give me some time :) i get a cubic in r3 that looks hard to solve

8. jimmyrep

r1 + r2 + r3 = a r1r2 + r1r3 + r2r3 = b r1r2r3 = c we need to get expressions for similar iidenties as the above replacing r1 by r1^4 etc in terms of a, b an c but that seems really dificult

9. jimmyrep

you need to find r1^4 + r2^4 + r3^4 in terms of the three identities and then substitute a, b and c and similarly for the other two identities - thats a daunting task

10. jamesm

tough one isn't it, jimmy?

11. jimmyrep

too right

12. jimmyrep

i'm trying to expand r1^4 + r2^4 + r3^4 - 4r1r2r3 (similar to what you do with a cubic) but its a real headache)

13. jimmyrep

best of luck! - must go - i have to take the wife shopping.

14. jamesm

no worries, have fun

15. jimmyrep

lol - u must be joking! - i'm spending money!

16. jamesm

i give up. do you have a solution abtrehearn?

17. myininaya

\[x^3-x^2*(a^4+R_a)+x(b^4-R_b)-c^4\]

18. myininaya

lol \[R_a\] means the remaining part of a^4 \[R_b\] means the remaining part of b^4

19. jamesm

the remaining part?

20. myininaya

a^4=r1^4+r2^4+r3^4+....+some other stuff there is a minus sign in front of this so i need to add the remaining stuff back

21. myininaya

so there should be a negative in front of R_a

22. myininaya

since there is already a negative in front of that parenthesis

23. jamesm

i'm not sure what R_a represents? can you find the remaining two coefficients in terms of a, b and c?

24. myininaya

R_a is the ramaining part of a^4 remember a^4=(r1+r2+r3)^4=r1^4+r2^4+r3^4+alot of other gibberish this gibberish is R_a

25. jamesm

yes, but the gibberish is in terms of r1, r2 and r3

26. myininaya

(r1+r2+r3)^4=r1^4+r2^4+r3^4+R(x)

27. abtrehearn

One thing that comes in handy is this identity: \[r _{1}^{3} + r _{2}^{3} + r _{3}^{3} - 3r _{1}r _{2}r _{3} = (r _{1} + r _{2} + r _{3})(r _{1}^{2} + r _{2}^{2} + r _{3}^{2} - r _{1} r _{2} - r _{1}r _{2}-r _{2}r _{3})\]

28. abtrehearn

\[= (r _{1} + r _{2} + r _{3})(r _{1}^{2} + r _{2}^{2} + r _{3}^{2} - r _{1}r _{2} - r _{1}r _{3} - r _{2}r _{3})\]

29. abtrehearn

\[= a(r _{1}^{2} + r _{2}^{2} + r _{3}^{2} - b)\] \[= r _{1}^{3} + r _{2}^{3} + r _{3}^{3} - 3c\] Another useful identity is \[\[(r _{1} + r _{2} + r _{3})^{2} = r _{1}^{2} + r _{2}^{2} + r _{3}^{2} + 2(r _{1}r _{2} + r _{1}r _{3} + r _{2}r _{3})\]

30. abtrehearn

This identity leads to \[a ^{2} = r _{1}^{2} + r _{2}^{2} + r _{3}^{2} + 2b.\]

31. abtrehearn

So a^2 - 2b can sub in for r1^2 + r2^2 + r3^2 in the cubic identity.

32. abtrehearn

If we have x^3 = a x^2 - b x + c for x in {r1, r2, r3}, the same is true for x^4 = a x^3 - b x^2 + c x = a(ax^2 - bx + c) + b x^2 + c x = (a^2 + b) x^2 + (-a b + c)