The 100 followers problem!

The given condition can also be expressed as $x^2z+y^2x+z^2y=0$ . Now, let $a=x^2z$ , $b=y^2x$ and $c=z^2y$ . Hence, $a+b+c=0$ and the required expression can be expressed as: $\dfrac{a^7+b^7+c^7}{(a^5+b^5+c^5)(a^2+b^2+c^2)}$

Let's use Newton's Sums. Let $S_1=a+b+c$ , $S_2=ab+ac+bc$ , $S_3=abc$ and $P_n=a^n+b^n+c^n$ .

$P_1=S_1=0 \\ P_2=P_1S_1-2S_2=-2S_2 \\ P_3=P_2S_1-P_1S_2+3S_3=3S_3 \\ P_4=P_3S_1-P_2S_2+P_1S_3=2S_2^2 \\ P_5=P_4S_1-P_3S_2+P_2S_3=-5S_2S_3 \\ P_6=P_5S_1-P_4S_2+P_3S_3=-2S_2^2+3S_3^2\\ P_7=P_6S_1-P_5S_2+P_4S_3=7S_2^2S_3$

So, the expression is now:

$\dfrac{P_7}{P_5P_2}=\dfrac{7S_2^2S_3}{(-5S_2S_3)(-2S_2)}=\dfrac{7S_2^2S_3}{10S_2^2S_3}=\boxed{\dfrac{7}{10}}$

Our final answer is then $7+10=\boxed{17}$ .