Complexity! (6)

Since,the answer is independent of x, take x=0. Hence, the value of expression is 7f(0)=7.

We note that $\alpha = e^\frac {i 2 \pi}7$ is the seventh root of unity. Therefore, $\alpha^6 + \alpha^5 + \alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1 = 0$ . Now we have:

$\begin{aligned} S & = f(x) + f(\alpha x) + f(\alpha^2 x) + f(\alpha^3 x) + f(\alpha^4 x) + f(\alpha^5 x) + f(\alpha^6 x) \\ & = \sum_{j=0}^6 f(\alpha^j x) \\ & = \sum_{j=0}^6 \left(1 + \sum_{k=1}^{20} a_k (\alpha^j x)^k \right) \\ & = 7 + \sum_{j=0}^6 \sum_{k=1}^{20} a_k (\alpha^j x)^k \\ & = 7 + \sum_{k=1}^{20} a_k x^k \sum_{j=0}^6 \alpha^{jk} \\ & = 7 + \sum_{k=1}^{20} a_k x^k \frac {\color{#3D99F6}{\alpha^ {7k}-1}}{\alpha^k-1} \quad \quad \small \color{#3D99F6}{\sum_{j=0}^6 \alpha^{jk} = 0 \text{ for } k \ne 7, 14 \text{ or } 7 \text{ for } k = 7, 14} \\ & = \boxed{7} \left(1 + a_7x^7 + a_{14}x^{14}\right) \end{aligned}$