A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what comes to mind is finding $(a + b + c + d)^3$ and subtracting whatever we don't need to get $a^3 + b^3 + c^3 + d^3$. However,
\begin{equation*}
(a+b+c+d)^3 = a^3+3a^2b+3a^2c+3a^2d+3ab^2+6abc+6abd+3ac^2+6acd+3ad^2+b^3+3b^2c+3b^2d+3bc^2+6bcd+3bd^2 + c^3 + 3c^2d + 3cd^2 + d^3
\end{equation*}
There is simply no good way to calculate $6abc + 6abd + 6acd + 6bcd $ without expanding everything.

You're not using the most important parts of the question: namely that $\xi$ is a fifth root of unity, and also that conveniently, all the coefficients are either $20$ and $13$.
Calculating $a^3 + b^3 + c^3 + d^3$ directly and collecting like coefficients, we have:
$$a^3 + b^3 + c^3 + d^3 = 20^3(\xi^6 + \xi^{12} + \xi^9 + \xi^3) + 3 \cdot 20^2 \cdot 13(\xi^5 + \xi^{10} + \xi^{10} + \xi^5)$$
$$+ 3 \cdot 20 \cdot 13^2 (\xi^4 + \xi^8 + \xi^{11} + \xi^{7}) + 13^3 (\xi^3 + \xi^6 + \xi^{12} + \xi^9)$$
and since $\xi^5 = 1$:
$$= 20^3 \left(\frac{1 - \xi^{15}}{1 - \xi^3} - 1 \right) + 3 \cdot 20^2 \cdot 13 \cdot4 +3 \cdot 20 \cdot 13^2 (\xi^4 + \xi^3 + \xi^1 + \xi^2) + 13^3 \left(\frac{1 - \xi^{15}}{1 - \xi^3} - 1 \right)$$
Can you continue from here?
