$\alpha, \beta, \gamma$ 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.
(Following up on my comment, too long to post as one.)
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 $\,r=0\,$.