<s>[INST] Finding $\sum_{k=0}^{6}a_{3k} $ where $(x^2-x+1)^{10}=\sum_{k=0}^{20}a_k x^k$ 
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
 [/INST] 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 $f(x)=(x^2-x+1)^{10}$.
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