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Source: AoPS

My attempt:

$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\implies144=50+2(ab+bc+ca) $$ $$\implies ab+bc+ca=47$$ and $$a^3+b^3+c^3-3abc=(a^2+b^2+c^2-(ab+bc+ca))(a+b+c)\implies216-3abc=12(50-47)$$$$\implies abc=(216-36)/3=60$$ So, $ab=\frac{60}{c}$

Now, $$a^3+b^3+c^3=216\implies a^3+b^3=216-c^3$$$$\implies(a+b)(a^2+b^2-ab)=216-c^3$$

In this equation, I substituted $a+b = 12 - c$, $a^2+b^2=50-c^2$, $ab=60/c$, and got a fourth degree polynomial in $c$ with complex roots which is terribly wrong.

Question: Where did I go wrong and how should I proceed?

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  • $\begingroup$ @Arthur we know $a < b < c$ which breaks the symmetry $\endgroup$ – Henno Brandsma Nov 28 '15 at 9:17
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Note that you have found the three elementary symmetric polynomials in the variables $a,b,c$, which determine a cubic polynomial whose roots are $a,b,c$. So $a,b,c$ are the solutions to \begin{equation*} \begin{aligned} &\mathrel{\phantom{=}} x^3 - (a+b+c)x^2 + (ab+bc+ac)x - abc \\ &= x^3 - 12x^2 + 47x - 60 \\ &= (x - 3)(x - 4)(x - 5). \end{aligned} \end{equation*} Given the condition $a < b < c$, we have $a = 3, b = 4, c = 5$, and so $a + 2b + 3c = 26$.

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