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Math Help - (αβ)(αγ)(βγ) = αβγ+αβγ+αβγ = [{α}/{αβγ}] + [{β}/{αβγ}] + [{γ}/{αβγ}]; No?

  1. #1
    PQR
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    (αβ)(αγ)(βγ) = αβγ+αβγ+αβγ = [{α}/{αβγ}] + [{β}/{αβγ}] + [{γ}/{αβγ}]; No?

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    Given that \alpha, \beta, \gamma are the roots of the equation x^3+x^2+4x-5 = 0, find the cubic equation whose roots are \alpha\beta, \alpha\gamma, \beta\gamma.
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    \alpha+\beta+\gamma = -1; \alpha\beta+\alpha\gamma+\beta\gamma = 4; \alpha\beta\gamma = 5.

     x^3-(\sum{a}_{1})x^2+(\sum{a}_{1}{a}_2)x-(\prod{a}_{1}{a}_{2}{a}_{3}) = 0

    \sum{a}_{1} = \alpha\beta+\alpha\gamma+\beta\gamma = 4

     \sum{a}_{1}{a}_2 = (\alpha\beta)(\alpha\gamma) (\beta\gamma)

    = \alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\bet  a\gamma^2

    = \frac{\alpha}{\alpha\beta\gamma}+\frac{\beta}{\alp  ha\beta\gamma} +\frac{\gamma}{\alpha\beta\gamma}

     = \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma} <br />

     <br />
= \frac{-1}{5}<br />

    \prod{a}_{1}{a}_{2}{a}_{3} = (\alpha\beta)(\alpha\gamma) (\beta\gamma) = \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = {5}^2 = 25 <br />

    Since  x^3-(\sum{a}_{1})x^2+(\sum{a}_{1}{a}_2)x-(\prod{a}_{1}{a}_{2}{a}_{3}) = 0

    the cubic equation is  <br />
x^3-4x^2-\frac{5}{1}x-25 = 0 <br />

    Well, it turns out that the sum of roots in pairs should have been -4, not  \frac{-1}{5}; and in checking, I'm hell sure that (\alpha\beta)(\alpha\gamma) (\beta\gamma) = \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Let x^3 + p x^2 + qx + r be the cubic you need to find.

    You know you have p = -4 and r = 5^2.

    Note that \alpha \beta \alpha \gamma + \alpha \beta \beta \gamma + \alpha \gamma \beta \gamma = (\alpha \beta \gamma)(\alpha + \beta + \gamma) which gives you your q.
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  3. #3
    PQR
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    Quote Originally Posted by Matt Westwood View Post
    Let x^3 + p x^2 + qx + r be the cubic you need to find.

    You know you have p = -4 and r = 5^2.

    Note that \alpha \beta \alpha \gamma + \alpha \beta \beta \gamma + \alpha \gamma \beta \gamma = (\alpha \beta \gamma)(\alpha + \beta + \gamma) which gives you your q.
    Thank you. I wonder why I haven't noticed that. But (\alpha \beta \gamma)(\alpha + \beta + \gamma) = (5)(-1) = -5 ; not -4.
    Last edited by PQR; September 4th 2009 at 12:15 AM.
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  4. #4
    PQR
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    Oh,  (\alpha\beta)(\alpha\gamma) (\beta\gamma) \neq \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}<br />
.
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  5. #5
    Senior Member pacman's Avatar
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    ah, nice observation
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