A smarter (not bashy) way to solve this roots of unity problem?

Question

(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.

Answer 1

They simplified exponents when defining $\xi,$ but you could also write the equations $$\begin{align*}a &= \xi(20\xi+13),\\b &= \xi^2(20\xi^2+13),\\c &= \xi^3(20\xi^3+13),\\d &= \xi^4(20\xi^4+13)\end{align*}$$ (actually, they swapped $c$ and $d$, but that doesn't change the problem). Consider the polynomial $$f(x) = [x(20x+13)]^3 = 20^3x^6+3\cdot20^2\cdot13x^5+3\cdot20\cdot13^2x^4+13^3x^3.$$ A roots of unity filter* using fifth roots of unity would give the sum of the coefficients of $x^0, x^5, x^{10},$ etc., which is just $3\cdot20^2\cdot 13 = 15600.$ But $\xi$ is a fifth root of unity! So, $$15600 = \frac{f(1)+f(\xi)+f(\xi^2)+f(\xi^3)+f(\xi^4)}{5} = \frac{33^3+a^3+b^3+c^3+d^3}{5}.$$ Rearranging gives $$a^3+b^3+c^3+d^3=78000-33^3=\boxed{42063}.$$

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*Say we have some polynomial $$P(x) = a_0x^0+a_1x^1+a_2x^2+\cdots+a_nx^n$$ Plugging in powers of the $k$th root of unity $\omega$ gives $$\sum_{j=0}^{k-1}P(\omega^j) = \sum_{i=0}^n a_i\sum_{j=0}^{k-1}(\omega^{j})^i.$$ Inspecting the inner summand, we find $$\sum_{j=0}^{k-1}(\omega^i)^j = k$$ if $\omega^i = 1$ (i.e. $i$ is a multiple of $k$), otherwise $$(\omega^i-1)\sum_{j=0}(\omega^i)^j = (\omega^{i})^k-1 = 0,$$ so it evaluates to zero.

Answer 2

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?

Answer 3

Let $\omega_1, \ldots, \omega_p$ be the $p$th roots of unity (including $1$), where $p$ is prime. Note that: $$\omega_1^k + \ldots + \omega_p^k = \begin{cases} p & \text{if } p \mid k \\ 0 & \text{otherwise.} \end{cases}$$ Why? The group of $p$th roots of unity are isomorphic to $\Bbb{Z} / p\Bbb{Z}$, and the map $x \mapsto x^k$ is equivalent to the map $x \mapsto kx$ on $\Bbb{Z} / 5 \Bbb{Z}$. It's well know that this homomorphism is trivial when $p \mid k$, and injective (and thus surjective) otherwise. So, when $p \not\mid k$, we are summing the $p$th roots of unity, which is $0$.

So, if we have a polynomial $q$, we can evaluate $q(\omega_1) + \ldots + q(\omega_p)$ relatively easily. We get: $$q(\omega_1) + \ldots + q(\omega_p) = p(q_0 + q_p + q_{2p} + \ldots),$$ where $q_i$ is the coefficient of $x^i$, and the sum on the right hand side is necessarily finite, due to the finite degree of $q$.

In your case, where $p = 5$, we have $a^3, b^3, c^3, d^3, (20 + 13)^3$ are (not necessarily respectively) $q(\omega_1), \ldots, q(\omega_5)$, where $$q(x) = (20x^2 + 13x)^3.$$ We only care about the coefficients of $x^5$ and $x^0$, the latter of which is clearly $0$, and the former is $3 \cdot 20^2 \cdot 13$. So, $$a^3 + b^3 + c^3 + d^3 + 33^3 = 15 \cdot 20^2 \cdot 13,$$ hence the answer is $15 \cdot 20^2 \cdot 13 - 33^3$

Answer 4

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 $$ \begin{align} a^3&=\left(20\xi^2+13\xi\right)^3&&=8000\xi+15600+10140\xi^4+2197\xi^3\\ b^3&=\varphi\!\left(a^3\right)&&=8000\xi^2+15600+10140\xi^3+2197\xi\\ c^3&=\varphi\circ\varphi\!\left(a^3\right)&&=8000\xi^4+15600+10140\xi+2197\xi^2\\ d^3&=\varphi\circ\varphi\circ\varphi\!\left(a^3\right)&&=8000\xi^3+15600+10140\xi^2+2197\xi^4\\\hline &\!\!\!\!\!a^3+b^3+c^3+d^3&&=-8000+62400-10140-2197\\ &&&=42063 \end{align} $$