Finding $\sum_{k=0}^{6}a_{3k} $ where $(x^2-x+1)^{10}=\sum_{k=0}^{20}a_k x^k$

Question

If $(x^2-x+1)^{10}=\displaystyle\sum_{k=0}^{20}a_k x^k$ then find the value of the expression $\displaystyle\sum_{k=0}^{6}a_{3k}$.

---

My Attempt

Since the subscripts on terms we are supposed to find involve a difference of $3$, substituting in $x=\omega$ and $x=-\omega$, where $\omega$ is a cube root of unity seems likely.

$$\begin{align} x=\omega &\implies(\omega ^2-\omega +1)^{10}=2^{10}\omega=\sum_{k=0}^{6}a_{3k}+\omega \sum_{k=0}^{6}a_{3k+1}+\omega ^2\sum_{k=0}^{6}a_{3k+2}\tag1 \\ x=-\omega &\implies(\omega ^2+\omega+1)^{10}=0 \ \ \ \ \ =\sum_{k=0}^{6}a_{3k}-\omega\sum_{k=0}^{6}a_{3k+1}+\omega^2\sum_{k=0}^{6}a_{3k+2} \tag2\end{align}$$ Adding $(1)$ and $(2)$ gets us rid of the term with $\omega$ but $\omega^2$ term just adds up to give an expression containing an extra $\sum_{k=0}^{6}a_{3k+2}$.

---

How can I proceed? Any hints are appreciated. Thanks

Answer 1

Use $x=1$, $x=\omega$ and $x=\omega^2$. You get $$3\sum_{k=0}^6a_{3k}=f(1)+f(\omega)+f(\omega^2)$$ where $f(x)=(x^2-x+1)^{10}$.

Answer 2

Using $\omega^2+ \omega+1=0 $ and $\omega ^3=1$, we get $$ \begin{aligned} f(\omega)&=\left(\omega^2-\omega+1\right)^{10}=(-2 \omega)^{10} =2^{10} \omega \\ f\left(\omega^2\right) & =\left(\omega^4-\omega^2+1\right)^{10} =\left(\omega-\omega^2+1\right)^{10} =\left(-2 \omega^2\right)^{10} =2^{10} \omega^{2} \end{aligned} $$ Hence we can conclude that $$ \begin{aligned} \sum_{k=0}^6 a_{3 k} & =\frac{1}{3}\left(1+2^{10} \omega+2^{10} \omega^{2}\right) \\ & =\frac{1}{3}\left[1+2^{10}\left(\omega+\omega^2\right)\right] \\ & =\frac{1}{3}\left(1-2^{10}\right) \\ & =-341 \end{aligned} $$