(αβ)(αγ)(βγ) = α²βγ+αβ²γ+αβγ² = [{α}/{αβγ}] + [{β}/{αβγ}] + [{γ}/{αβγ}]; No?

**PQR** · September 3rd 2009, 10:44 PM

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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} $

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

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

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

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 $(\alpha\beta)(\alpha\gamma) (\beta\gamma) = \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}.$

**Matt Westwood** · September 3rd 2009, 10:59 PM

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 $q$ .

**PQR** · September 3rd 2009, 11:23 PM

Originally Posted by Matt Westwood

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 $-4$ .

**PQR** · September 4th 2009, 12:02 AM

Oh, $(\alpha\beta)(\alpha\gamma) (\beta\gamma) \neq \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma} $ .

**pacman** · September 4th 2009, 08:59 PM

ah, nice observation

Thread: (αβ)(αγ)(βγ) = α²βγ+αβ²γ+αβγ² = [{α}/{αβγ}] + [{β}/{αβγ}] + [{γ}/{αβγ}]; No?

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