An algebra problem by A Former Brilliant Member

Using Newton's sums method and let $P_n = a^n + b^n + c^n$ , where $n$ is a positive integer, $S_1 = a+b+c = 0$ , $S_2 = ab+bc+ca$ , and $S_3 = abc$ . Then, we have:

\(\begin{array} {} P_1 = S_1 & = 0 & = 0 \\ P_2 = S_1P_1 - 2S_2 & = 0 - 2 S_2 & = - 2S_2 \\ P_3 = S_1P_2 - S_2P_1 + 3S_3 & = 0 - 0 + 3S_3 & = 3S_3 \\ P_4 = S_1P_3 - S_2P_2 + S_3P_1 & = 0 - S_2(-2S_2) + 0 & = 2S_2^2 \\ P_5 = S_1P_4 - S_2P_3 + S_3P_2 & = 0 - S_2(3S_3) + S_3(-2S_2) & = -5S_2S_3 \end{array} \)

$\implies \dfrac {(a^3 + b^3 + c^3)^2(a^4 + b^4 + c^4)}{(a^5 + b^5 + c^5)^2} = \dfrac {P_3^2P_4}{P_5^2} = \dfrac {(3S_3)^2(2S_2^2)}{(-5S_2S_3)^2} = \dfrac {18}{25}$

$\implies m + n = 18+25 = \boxed{43}$

Using Newton sum.

Firstly finding $(a^3+b^3+c^3)$ .

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2)-(ab-bc-ca)(a+b+c)+3abc$
$a^3+b^3+c^3=3abc$
$\Rightarrow (a^3+b^3+c^3)^2=\boxed{9\color{#D61F06}{(abc)^2}}$

Secondly finding $(a^4+b^4+c^4)$ .

$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2[(ab)^2+(bc)^2+(ca)^2]$
$a^4+b^4+c^4=[(a+b+c)^2-2(ab+bc+ca)]-2[(ab)^2+(bc)^2+(ca)^2]$
$\Rightarrow a^4+b^4+c^4=\boxed{2\color{#20A900}{\left[(ab)^2+(bc)^2+(ca)^2\right]}}$

Thirdly finding $(a^5+b^5+c^5)$ .

$a^5+b^5+c^5=(a+b+c)(a^4+b^4+c^4)-(ab+bc+ca)(a^3+b^3+c^3)+abc(a^2+b^2+c^2)$
$a^5+b^5+c^5=(ab+bc+ca)(-5abc)$
$\Rightarrow (a^5+b^5+c^5)^2=(ab+bc+ca)^2(-5abc)^2=\boxed{\color{#D61F06}{\left[(ab)^2+(bc)^2+(ca)^2\right]}(-5\color{#D61F06}{abc})^2}$

$\implies \dfrac{(a^3+b^3+c^3)^2(a^4+b^4+c^4)}{(a^5+b^5+c^5)^2}$

$\Rightarrow \dfrac{9\color{#D61F06}{(abc)^2}×2\color{#20A900}{\left[(ab)^2+(bc)^2+(ca)^2\right]}}{\color{#20A900}{\left[(ab)^2+(bc)^2+(ca)^2\right]}×25\color{#D61F06}{(abc)^2}}$

$\Rightarrow \dfrac{18}{25}=\dfrac{m}{n}$

$\Rightarrow m+n=18+25=\boxed{43}$