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Thread: (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 $\displaystyle \alpha$, $\displaystyle \beta$, $\displaystyle \gamma $ are the roots of the equation $\displaystyle x^3+x^2+4x-5 = 0,$ find the cubic equation whose roots are $\displaystyle \alpha\beta,$ $\displaystyle \alpha\gamma,$ $\displaystyle \beta\gamma.$
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    $\displaystyle \alpha+\beta+\gamma = -1$; $\displaystyle \alpha\beta+\alpha\gamma+\beta\gamma = 4$; $\displaystyle \alpha\beta\gamma = 5.$

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

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

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

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

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

    $\displaystyle = \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}
    $

    $\displaystyle
    = \frac{-1}{5}
    $

    $\displaystyle \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
    $

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

    the cubic equation is $\displaystyle
    x^3-4x^2-\frac{5}{1}x-25 = 0
    $

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

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

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

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

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