Sum of powers of a field's element equals zero

Question

Assume that we have a finite field $\mathbb{F}_q$ and an element $a\in\mathbb{F}_q$ s.t. $\mathrm{ord}(a)=9$. I want to prove that $$1+a^{3}+a^{-3}+1+a^{3}+a^{-3}+1+a^{3}+a^{-3}=0$$ I know that this should hold by some coding theory property but I cannot prove it.

Answer 1

If $a$ has order $9$, then $a^3$ has order $3$, and so $1,a^3, a^{-3}$ are all different, and each of them is a root of $x^3-1$.

In a field, a nonzero polynomial cannot have more roots than its degree, so $1+a^3+a^{-3}$ is the sum of the three roots of $x^3-1$. But the sum of the roots of a monic polynomial (that has the maximal number of roots) is exactly minus the next-to-highest coefficient, which is $0$ in this case.

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Alternatively, rewrite your goal by multiplying by powers of $a^9$ to $$ 1+a^3+a^6+\cdots+a^{24}=0 $$ and apply the formula for a finite geometric series, giving $$ \frac{a^{27}-1}{a^3-1} = 0 $$ But the numerator here is $a^{27}-1=(a^9)^3-1=1^3-1=0$.