<s>[INST] Finding solutions to a system of equations using elementary symmetric polynomials Find the value of a, b, and c given:
$a^2 + b^2 + c^2 = 129$
$ab + ac + bc = -4$
$(a^2)(b^2) + (a^2)(c^2) + (b^2)(c^2)$ = 984
I attempted this problem using elementary symmetrical polynomials and found that the second equation can be written as $e_2(a, b, c)$. However, the squares in the other equations made me quite confused and I was unable to write them in the previous from. If I was able to write it in the form of elementary symmetrical polynomials, I should have been able to use simple substitution to find the value of $e_1, e_2$, and $e_3$. This then would have allowed me to form a cubic equation, where I could find the solutions by solving it.
However, as I mentioned before, I'm a little bit stuck on the part where I need to find the values of $e_1, e_2, e_3$. Any help would be extremely appreciated :) Also, is there any other way to solve for a, b, and c using elementary symmetric polynomials and substitution, but not cubic equations?
Update:
Thanks to @Gerry_Myerson, I've worked out that the first equation is:
$(e_1)^2 - 2e_2 = 129$
and the second equation is:
$e_2 = -4$
However, I still don't know how to work out the last one.
 [/INST] Let $P(t) = (t-a)(t-b)(t-c) = t^3-e_1t^2+e_2t-e_3$. 
Trivially, $e_1 = a+b+c$, $e_2 = ab+bc+ca = -4$, and $e_3 = abc$. 
As noted in the comments, $129 = a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) = e_1^2-2e_2 = e_1^2+8$, so $e_1^2 = 121$, and thus, $e_1 = \pm 11$. 
Next, note that $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} = \left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2 - 2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right) = \left(\dfrac{ab+bc+ca}{abc}\right)^2 - 2\dfrac{a+b+c}{abc} = \left(\dfrac{e_2}{e_3}\right)^2 - \dfrac{2e_1}{e_3}$
Multiplying both sides by $a^2b^2c^2 = e_3^2$ yields
$984 = a^2b^2+b^2c^2+c^2a^2 = e_2^2-2e_1e_3 = (-4)^2-2(\pm 11)e_3$, and thus, $e_3 = \mp44$. 
So, $a,b,c$ are the roots of either $t^3-11t^2-4t+44 = 0$ or $t^3+11t^2-4t-44 = 0$
Solving these equations yeilds $(a,b,c) = (2,-2,11)$ and $(2,-2,-11)$ and permutations as solutions.
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