Possible formula for $\sum_{} (\alpha\beta)^2$ and $\sum_{} \alpha\beta(\alpha+\beta)$

Question

Given a quartic equation in the form $$ax^4+bx^3+cx^2+dx+e=0$$ and its roos $\alpha, \beta, \gamma$ and $\delta$, can anyone please help me find a formula for $$\sum_{} (\alpha\beta)^2 \quad \text{and} \quad \sum_{} \alpha\beta(\alpha+\beta)\;?$$

Answer 1

Since we have a quartic equation, we see that $a\neq0.$

Thus, the Viete's theorem we obtain: $$\frac{1}{4}\sum_{sym}\alpha^2\beta^2=\left(\frac{1}{4}\sum_{sym}\alpha\beta\right)^2-2\sum_{cyc}\alpha\sum_{cyc}\alpha\beta\gamma+\prod_{cyc}\alpha=$$ $$=\left(\frac{c}{a}\right)^2-2\left(-\frac{b}{a}\right)\left(-\frac{d}{a}\right)+2\cdot\frac{e}{a}=\frac{c^2-2bd+2ae}{a^2}.$$ $$\sum_{cyc}\alpha^2(\beta+\gamma+\delta)=\sum_{cyc}\alpha\cdot\frac{1}{4}\sum_{sym}\alpha\beta-3\sum_{cyc}\alpha\beta\gamma=$$ $$=-\frac{b}{a}\cdot\frac{c}{a}-3\left(-\frac{d}{a}\right)=\frac{3ad-bc}{a^2}.$$

PS. $$\sum_{sym}\alpha\beta=4(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta).$$ $$\sum_{cyc}\alpha^2(\beta+\gamma+\delta)=\alpha^2(\beta+\gamma+\delta)+\beta^2(\gamma+\delta+\alpha)+\gamma^2(\delta+\alpha+\beta)+\delta^2(\alpha+\beta+\gamma).$$ By the way, $$\sum_{sym}\alpha^2\beta=2\left(\alpha^2(\beta+\gamma+\delta)+\beta^2(\gamma+\delta+\alpha)+\gamma^2(\delta+\alpha+\beta)+\delta^2(\alpha+\beta+\gamma)\right).$$