Sequence and series

Relevant wiki: Newton's Identities

Using Newton's sums or Newton's identities and let $P_n = a^n + b^n + c^n$ , where $n$ is a positive integer, $S_1 = P_1 = a+b+c = 0$ , $S_2 = ab+bc+ca$ , and $S_3 = abc$ . Then we have:

$\begin{aligned} P_7 & = a^7+b^7+c^7 \\ & = {\color{#3D99F6}S_1}P_6 - S_2P_5 + S_3P_4 & \small \color{#3D99F6} \text{Recall }S_1 = 0 \\ & = {\color{#3D99F6}0} - S_2(-S_2P_3+S_3P_2) + S_3(-S_2P_2 + S_3{\color{#3D99F6}P_1}) & \small \color{#3D99F6} \text{also }P_1 = 0 \\ & = S_2^2{\color{#3D99F6}P_3} - 2S_2S_3{\color{#D61F06}P_2} \\ & = S_2^2{\color{#3D99F6}(S_1P_2-S_2P_1 + 3S_3)} - 2S_2S_3{\color{#D61F06}(S_1P_1-2S_2)} \\ & = S_2^2{\color{#3D99F6}(3S_3)} - 2S_2S_3{\color{#D61F06}(-2S_2)} & \small \color{#3D99F6} \implies P_3 = 3S_3 \\ & = 7S_2^2S_3 & \small \color{#D61F06} \implies P_2 = - 2S_2 \end{aligned}$

Therefore, $\dfrac {7(a^2+b^2+c^2)^2(a^3+b^3+c^3)}{a^7+b^7+c^7} = \dfrac {7{\color{#D61F06}P_2}^2\color{#3D99F6}P_3}{P_7} = \dfrac {7({\color{#D61F06}-2S_2})^2 ({\color{#3D99F6}3S_3})}{7S_2^2S_3} = \boxed{12}$ .

Just set $a,b,c$ as three roots of unity IE $\omega, {\omega}^2, 1$ and you are done.

We know that, $a^3 + b^3 + c^3 = 3abc$ when $a + b + c =0 \Rightarrow c = -a - b.$

$(a^2 + b^2 + c^2)^2 = ((a + b + c) ^2 - 2(ab + bc + ca) )^2 =( (0)^2 - 2(ab + bc + ca))^2 = 4(a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc( a + b + c) ) =4 ( a^2 b^2 + b^2 c^2 + c^2a^2 )$

Using Binomial Theorem ,

$a^7 + b^7 + c^7 = a^7 + b^7 + (-a - b)^7 =- 7ab(a^5 + b^5 + 3ab( a^3 + b^3) + 5a^2 b^2 (a + b)) = D$

Above expression becomes , $\dfrac {7\times 4(a^2 b^2 + b^2 c^2 + c^2 a^2)(3abc)}{- 7ab(a^5 + b^5 + 3ab( a^3 + b^3) + 5a^2 b^2 (a + b)) } = \dfrac ND$

$N = (12\times 7)(a^2 b^2 + b^2 (-a -b) ^2 + (-a -b) ^2 a^2 )(ab(-a -b)) = (12\times (-7ab (a^5 + b^5 + 2ab(a^3 + b^3 + ab^2 + a^2 b) + 4a^2 b^2(a + b) )) = 12 (D)$

Therefore, $\dfrac ND = \dfrac {12 \times D}{D} = \boxed {12}$

The last step might be a bit confusing because I myself am a bit confused. Please let me know if there is a easy way of doing this.

I set random values for $a$ , $b$ , and $c$ , such that $a+b+c=0$ (except the case where $a=b=c=0$ , because the answer would be indeterminate). I assumed that no matter what the specific values of $a$ , $b$ , and $c$ were, they would yield the same answer, otherwise there would be many different answers, possibly infinite. So I set $a=1$ , $b=1$ , and $c=-2$ . Plug them into the equation and you will get $\frac{7(a^{2}+b^{2}+c^{2})^{2}(a^{3}+b^{3}+c^{3})}{(a^{7}+b^{7}+c^{7})}=\frac{7((1)^{2}+(1)^{2}+(-2)^{2})^{2}((1)^{3}+(1)^{3}+(-2)^{3})}{((1)^{7}+(1)^{7}+(-2)^{7})}=12$ .