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A quartic polynomial with roots $\alpha$, $\beta$, $\gamma$, $\delta$ has $\sum \alpha = a$, $\sum \alpha\beta = b$, $\sum\alpha \beta \gamma = c$ and $\sum \alpha\beta\gamma\delta = d$.

Find the value of $\sum (\alpha \beta)^2$ in terms of $a, b, c$ and $d$.

Is there a quick way to do this or do you need to consider

$( (\alpha\beta) + (\beta\gamma) + (\gamma\alpha) + (\gamma\delta) + (\alpha\delta) + (\beta\delta) )^2$

then expand and work through the long algebra?

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  • $\begingroup$ Sorry this is the notation used in my course. E.g. $\sum \alpha \beta$ means the sum of the products of two of the roots. =(αβ)+(βγ)+(γα)+(γδ)+(αδ)+(βδ) $\endgroup$ – MGross Jan 17 '18 at 19:57
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One way is to do as you suggest in the question ... calculate \begin{eqnarray*} (\sum \alpha \beta )^2 = \sum \alpha^2 \beta ^2 + 2\sum \alpha^2\beta \gamma +6 \alpha \beta \gamma \delta \\ (\sum \alpha ) ( \sum \alpha \beta \gamma ) = \sum \alpha^2\beta \gamma +4 \alpha \beta \gamma \delta \\ \end{eqnarray*}

An alternative is to rearrange the polynomial to $x^4+b x^2+d=ax^3+cx$, square this \begin{eqnarray*} x^8+(2b-a^2)x^6+(\color{red}{2d+b^2-2ac})x^4 +(2d-c^2)x^2+d^2=0 \end{eqnarray*} The roots of this quartic in $x^2$ will be the squares of the roots of the previous quartic so we have \begin{eqnarray*} \sum \alpha^2 \beta ^2 = 2d+b^2-2ac. \end{eqnarray*}

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  • $\begingroup$ Thanks. I should have realised that I need to consider the polynomial with roots $\alpha^2, \beta^2, \gamma^2$ and $\delta^2$. I'm more used to doing this kind of thing using substitution $u=\sqrt{x}$ but it's just as fast. $\endgroup$ – MGross Jan 17 '18 at 20:50
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$$\frac{1}{4}\sum_{sym}\alpha^2\beta^2=\left(\frac{1}{4}\sum_{sym}\alpha\beta\right)^2-2\sum_{cyc}\alpha\sum_{cyc}\alpha\beta\gamma+2\alpha\beta\gamma\delta=b^2-2ac+2d.$$ I used $(x+y+z+t+u+v+w)^2=x^2+y^2+....$

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