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?


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.

  • 1
    $\begingroup$ Note that (a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)$. $\endgroup$ – Gerry Myerson Jun 10 at 0:19
  • $\begingroup$ The 3rd equation is currently missing an equal sign and a right side. $\endgroup$ – JimmyK4542 Jun 10 at 0:24
  • $\begingroup$ Thanks for letting me know - I've fixed it now $\endgroup$ – Alexander B Jun 10 at 0:28

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.

  • $\begingroup$ So would there be 12 sets of solutions? $\endgroup$ – Alexander B Jun 10 at 1:01
  • $\begingroup$ Yes, I believe so. $\endgroup$ – JimmyK4542 Jun 10 at 1:03

\begin{align} a^2 + b^2 + c^2 &= 129 \tag{1}\label{1} ,\\ ab + bc + ac &= -4 \tag{2}\label{2} ,\\ a^2b^2 + b^2c^2 + a^2c^2&=984 \tag{3}\label{3} . \end{align}

\begin{align} \eqref{2}^2:\quad a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c) &=16 ,\\ abc(a+b+c)&=-484 ,\\ \eqref{1}+2\cdot\eqref{2}:\quad (a+b+c)^2&=121 .\\ a+b+c&=\pm 11 ,\\ abc&=-\frac{484}{a+b+c} =\mp 44 , \end{align}

and we have two cubics \begin{align} x^3-11x^2-4x+44 ,\\ y^3+11y^2-4x-44 \end{align}

with roots $(-2, 2, 11)$ and $(-2, 2, -11)$.


Suposse that $a$, $b$ and $c$ are roots of a polynomial of degree 3. $$mx^3+nx^2+px+q$$ Then by Vieta's formulas $$a+b+c=\tfrac{-n}{m}, ab+ac+bc=\tfrac{p}{m}, abc=\tfrac{-q}{m}$$. Note that you can found

  • $a+b+c$, if expand $(a+b+c)^2$

  • $abc$, if expand $(ab+ac+bc)^2$


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