Are you making its root?

$\begin{aligned} \sum_{a, \ b, \ c} \frac {a^6-1}{a-1} & = \sum_{a, \ b, \ c} \frac {(a^3-1)(a^3+1)}{a-1} \\ & = \sum_{a, \ b, \ c} \frac {(a-1)(a^2+a+1)(\color{#3D99F6}{a^3+1})}{a-1} & \small \color{#3D99F6}{\text{Note 1}} \\ & = \sum_{a, \ b, \ c} (a^2+a+1)(\color{#3D99F6}{8-3a}) \\ & = \sum_{a, \ b, \ c} (5a^2+5a+8 - 3\color{#3D99F6}{a^3}) \\ & = \sum_{a, \ b, \ c} (5a^2+5a+8 - 3\color{#3D99F6}{7-3a}) \\ & = \sum_{a, \ b, \ c} (5a^2+14a-13) \\ & = 5(\color{#3D99F6}{a^2+b^2+c^2})+14(\color{#D61F06}{a+b+c})-3(13) & \small \color{#3D99F6}{\text{Note 2}} \\ & = 5(\color{#3D99F6}{0^2-2(3)})+14(\color{#D61F06}{0})-39 \\ & = \boxed{-69} \end{aligned}$

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Notes:

1. Since $a$ is a root of $x^3+3x-7$ , then we have $a^3+3a-7=0$ , $\implies a^3 = 7-3a$ ; the same for $b$ and $c$ .
2. By Vieta's formula , we have $\color{#D61F06}{a+b+c = 0}$ and $ab+bc+ca = 3$ . Using the identity $\color{#3D99F6}{a^2+b^2+c^2 = (a+b+c)^2 - 2(ab_bc+ca)}$ $\color{#3D99F6}{= 0^2 - 2(3) = -6}$ .