<s>[INST] Let $a$, $b$, and $c$ be positive real numbers with $aSource: 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?
 [/INST] 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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