jmc

Employing the elementary symmetric polynomials ($s_1={\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4=-2$, $s_2={\alpha }_1{\alpha }_2+{\alpha }_1{\alpha }_3+{\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3+{\alpha }_2{\alpha }_4+{\alpha }_3{\alpha }_4=0$, $s_3={\alpha }_1{\alpha }_2{\alpha }_3+{\alpha }_2{\alpha }_3{\alpha }_4+{\alpha }_3{\alpha }_4{\alpha }_1+{\alpha }_4{\alpha }_1{\alpha }_2=0$, and $s_4={\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4=2$) we consider the polynomial $$P(x)=(x-({\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4))(x-({\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4))(x-({\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3))$$Because $P$ is symmetric with respect to ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$, we can express the coefficients of its expanded form in terms of the elementary symmetric polynomials. We compute \begin{eqnarray} P(x) & = & x^3 - s_2x^2 + (s_3s_1-4s_4)x + (-s_3^2-s_4s_1^2+s_4s_2) \\ & = & x^3 - 8x - 8 \\ & = & (x+2)(x^2-2x-4) \end{eqnarray}The roots of $P(x)$ are $-2$ and $1\pm \sqrt{5}$, so the answer is $\boxed{\{1\pm \sqrt{5},-2\}}.$

$\text{Remarks:}$ It is easy to find the coefficients of $x^2$ and $x$ by expansion, and the constant term can be computed without the complete expansion and decomposition of $({\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4)({\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4)({\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3)$ by noting that the only nonzero 6th degree expressions in $s_1,s_2,s_3,$ and $s_4$ are $s_1^6$ and $s_4{s}_1^2$. The general polynomial $P$ constructed here is called the cubic resolvent and arises in Galois theory.