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abtrehearn

  • 3 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?

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  1. jamesm
    • 3 years ago
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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

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

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

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

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

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

  7. jamesm
    • 3 years ago
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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

  8. jimmyrep
    • 3 years ago
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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

  9. jimmyrep
    • 3 years ago
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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

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

  11. jimmyrep
    • 3 years ago
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    too right

  12. jimmyrep
    • 3 years ago
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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)

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

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

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

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

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

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

  19. jamesm
    • 3 years ago
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    the remaining part?

  20. myininaya
    • 3 years ago
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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

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

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

  23. jamesm
    • 3 years ago
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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?

  24. myininaya
    • 3 years ago
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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

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

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

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

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

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

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

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

  32. abtrehearn
    • 3 years ago
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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)

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