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abtrehearn Group Title

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?

  • 3 years ago
  • 3 years ago

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  1. jamesm Group Title
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    (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

    • 3 years ago
  2. jamesm Group Title
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    struggling to see how to express the other terms in terms of a, b and c

    • 3 years ago
  3. abtrehearn Group Title
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    Looks good so far.

    • 3 years ago
  4. myininaya Group Title
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    isn't the answer just P(x)=(x-r1^4)(x-r2^4)(x-r3^4)

    • 3 years ago
  5. abtrehearn Group Title
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    That's right.

    • 3 years ago
  6. abtrehearn Group Title
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    The coefficients of the polynomial we're looking for are to be expressed in terms of a, b, and c.

    • 3 years ago
  7. jamesm Group Title
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    requires an algebraic trick that i can not see...give me some time :) i get a cubic in r3 that looks hard to solve

    • 3 years ago
  8. jimmyrep Group Title
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    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

    • 3 years ago
  9. jimmyrep Group Title
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    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

    • 3 years ago
  10. jamesm Group Title
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    tough one isn't it, jimmy?

    • 3 years ago
  11. jimmyrep Group Title
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    too right

    • 3 years ago
  12. jimmyrep Group Title
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    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)

    • 3 years ago
  13. jimmyrep Group Title
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    best of luck! - must go - i have to take the wife shopping.

    • 3 years ago
  14. jamesm Group Title
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    no worries, have fun

    • 3 years ago
  15. jimmyrep Group Title
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    lol - u must be joking! - i'm spending money!

    • 3 years ago
  16. jamesm Group Title
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    i give up. do you have a solution abtrehearn?

    • 3 years ago
  17. myininaya Group Title
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    \[x^3-x^2*(a^4+R_a)+x(b^4-R_b)-c^4\]

    • 3 years ago
  18. myininaya Group Title
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    lol \[R_a\] means the remaining part of a^4 \[R_b\] means the remaining part of b^4

    • 3 years ago
  19. jamesm Group Title
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    the remaining part?

    • 3 years ago
  20. myininaya Group Title
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    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

    • 3 years ago
  21. myininaya Group Title
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    so there should be a negative in front of R_a

    • 3 years ago
  22. myininaya Group Title
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    since there is already a negative in front of that parenthesis

    • 3 years ago
  23. jamesm Group Title
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    i'm not sure what R_a represents? can you find the remaining two coefficients in terms of a, b and c?

    • 3 years ago
  24. myininaya Group Title
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    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

    • 3 years ago
  25. jamesm Group Title
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    yes, but the gibberish is in terms of r1, r2 and r3

    • 3 years ago
  26. myininaya Group Title
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    (r1+r2+r3)^4=r1^4+r2^4+r3^4+R(x)

    • 3 years ago
  27. abtrehearn Group Title
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    One thing that comes in handy is this identity: \[r _{1}^{3} + r _{2}^{3} + r _{3}^{3} - 3r _{1}r _{2}r _{3} = (r _{1} + r _{2} + r _{3})(r _{1}^{2} + r _{2}^{2} + r _{3}^{2} - r _{1} r _{2} - r _{1}r _{2}-r _{2}r _{3})\]

    • 3 years ago
  28. abtrehearn Group Title
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    \[= (r _{1} + r _{2} + r _{3})(r _{1}^{2} + r _{2}^{2} + r _{3}^{2} - r _{1}r _{2} - r _{1}r _{3} - r _{2}r _{3})\]

    • 3 years ago
  29. abtrehearn Group Title
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    \[= a(r _{1}^{2} + r _{2}^{2} + r _{3}^{2} - b)\] \[= r _{1}^{3} + r _{2}^{3} + r _{3}^{3} - 3c\] Another useful identity is \[\[(r _{1} + r _{2} + r _{3})^{2} = r _{1}^{2} + r _{2}^{2} + r _{3}^{2} + 2(r _{1}r _{2} + r _{1}r _{3} + r _{2}r _{3})\]

    • 3 years ago
  30. abtrehearn Group Title
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    This identity leads to \[a ^{2} = r _{1}^{2} + r _{2}^{2} + r _{3}^{2} + 2b.\]

    • 3 years ago
  31. abtrehearn Group Title
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    So a^2 - 2b can sub in for r1^2 + r2^2 + r3^2 in the cubic identity.

    • 3 years ago
  32. abtrehearn Group Title
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    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)

    • 3 years ago
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