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?

  • $\begingroup$ @Arthur we know $a < b < c$ which breaks the symmetry $\endgroup$ – Henno Brandsma Nov 28 '15 at 9:17

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 $a + 2b + 3c = 26$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.