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.

Let $\varphi:\mathbb{Q}[\xi]\to\mathbb{Q}[\xi]$ be the field homomorphism which is the identity on $\mathbb{Q}$ and $\varphi(\xi)=\xi^2$.
Since $\xi+\xi^2+\xi^3+\xi^4=-1$, we have
$$
\begin{align}
a^3&=\left(20\xi^2+13\xi\right)^3&&=8000\xi+15600+10140\xi^4+2197\xi^3\\
b^3&=\varphi\!\left(a^3\right)&&=8000\xi^2+15600+10140\xi^3+2197\xi\\
c^3&=\varphi\circ\varphi\!\left(a^3\right)&&=8000\xi^4+15600+10140\xi+2197\xi^2\\
d^3&=\varphi\circ\varphi\circ\varphi\!\left(a^3\right)&&=8000\xi^3+15600+10140\xi^2+2197\xi^4\\\hline
&\!\!\!\!\!a^3+b^3+c^3+d^3&&=-8000+62400-10140-2197\\
&&&=42063
\end{align}
$$
