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Please no complete solutions, ONLY HINTS REQUESTED!

The complex numbers $\alpha_1$, $\alpha_2$, $\alpha_3$, and $\alpha_4$ are the four distinct roots of the equation $x^4+2x^3+2=0$. Determine the unordered set $\{\alpha_1\alpha_2+\alpha_3\alpha_4,\alpha_1\alpha_3+\alpha_2\alpha_4,\alpha_1\alpha_4+\alpha_2\alpha_3\}.$

Hints please. By Vieta's formulas, shows $\sigma_{2} = \sum_{\text{sym}} ab = a_{2} = 0$?

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  • $\begingroup$ You may want to (or avoid - because spoilers - to want to) look in the wikipedia page on quartic equations. These quantities play a role in a more number-theoretic treatment of their algebraic solution, in fact they are the solutions of the auxillary cubic polynomial that is part of the solution process. $\endgroup$ – Lutz Lehmann Oct 29 '15 at 19:45
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    $\begingroup$ Hi, here is a hint: Consider the polynomial $Q(x)$ with roots $\alpha_1\alpha_2+\alpha_3\alpha_4, \alpha_1\alpha_3+\alpha_2\alpha_4,\alpha_1\alpha_4+\alpha_2\alpha_3$. What can we say about coefficiants of $Q(x)$? $\endgroup$ – Max Oct 29 '15 at 23:12
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Hint: Construct the cubic whose roots are elements of the desired unordered set using Vieta's relations. Then solve it.


Getting into a bit more detail, as it seems that hint isn't sufficient.

Note $s_1=\sum \alpha_1 = -2, \;s_2=\sum \alpha_1 \alpha_2 = 0, \; s_3 =\sum \alpha_1\alpha_2\alpha_3=0, \; s_4=\prod \alpha_1 = 2$ from the given quartic.

For the new cubic, we have
sum of roots $= s_2 = 0$
sum of product of roots taken two at a time $= s_1s_3-4s_4=-8$
product of roots $=s_4s_1^2-4s_4s_2+s_3^2 =8$
Thus the cubic we seek is $x^3-8x-8=(x+2)(x^2-2x-4)$

I suppose finishing that is easy...

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