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$\alpha, \beta, \gamma$ are the roots of equation $x^3+px^2+qx+r=0$ . Find $$\sum{\frac{\alpha^2 + \beta^2}{\alpha+\beta}}.$$

I can solve the symmetric expression by : $$\frac{\alpha^2 + \beta^2}{\alpha+\beta} = (\alpha+\beta) -\frac{2\alpha \beta}{(\alpha+\beta) }.$$

Then, advance forward. But I wanted to know, if there is any other simpler method. I've seen such symmetric expressions solved by multiplying the inner terms and simpler terms pop out . But here I cannot...

Please help. Thanks.

Leucippus
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Entrepreneur
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  • You could simplify the second term using that $,\alpha\beta=-r / \gamma,$ and $,\alpha+\beta=-p-\gamma,$ (btw, there is no square in the denominator), then decompose $,1 / ( \gamma(p+\gamma) ),$ into partial fractions. – dxiv Sep 20 '18 at 18:17
  • @dxiv... I know that method... But, I wanted to know if there are any simplification of the expression. Like - in $\sum(\alpha-\beta) ^2$ , we can apply formula and break the term and easily get $2\sum {\alpha}^2 -2\sum \alpha \beta$ ... Anything like that applicable here? – Entrepreneur Sep 21 '18 at 04:12
  • Elaborated some more on my comment in an answer, since it got too long. But I don't know that you'll find an entirely painless way to work it out. – dxiv Sep 21 '18 at 04:57

1 Answers1

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(Following up on my comment, too long to post as one.)

$\frac{(\alpha) ^2+(\beta) ^2}{\alpha+\beta} = (\alpha+\beta) -\frac{2\alpha \beta}{(\alpha+\beta) }$

Also using that $\,\alpha\beta=-r /\gamma\,$ and $\,\alpha+\beta=-p-\gamma\,$:

$$ A = \sum \frac{\alpha ^2+\beta ^2}{\alpha+\beta} = \sum \left((\alpha+\beta) -\frac{2\alpha \beta}{\alpha+\beta} \right) = 2 \underbrace{\,\sum \alpha\,}_{=\,-p} + 2r \underbrace{\,\sum \frac{1}{\alpha(p+\alpha)}\,}_{= \,B} $$

Since $\,\dfrac{1}{\alpha(p+\alpha)}=\dfrac{1}{p}\left(\dfrac{1}{\alpha}-\dfrac{1}{p+\alpha}\right)\,$:

$$ B = \frac{1}{p}\left(\underbrace{\,\sum \frac{1}{\alpha}\,}_{=C} - \underbrace{\,\sum \frac{1}{p+\alpha}\,}_{= D}\right) $$

$\dfrac{1}{\alpha}\,$ are the roots of the reciprocal polynomial $\,x^3 f\left(\frac{1}{x}\right) = 1+px+qx^2+rx^3\,$, so:

$$ C = \sum \frac{1}{\alpha} = -\frac{q}{r} $$

$p+\alpha$ are the roots of $\,f(x-p) = x^3 - 2 p x^2 + (p^2 + q) x - p q + r\,$, so:

$$ D = \sum \frac{1}{p+\alpha}=-\frac{p^2 + q}{- p q + r} $$

What's left is to piece it all back together, and also cover the special cases like $\,r=0\,$.

dxiv
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