Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$. Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$.
So I am really stuck on this one.. I immediately defeated when I saw that I need to compute  $\binom {22} {a,b,c}$ where $a,b,c$ is the coefficients of $x^2,-x,1$. So what should be the next step I need to take? 

Using that $(a+b+c)^3\equiv a^3+b^3+c^3\pmod 3$ repeatedly, we have, in mod $3$,
$$\begin{align}&(x^2-x+1)^{22}\\\\&=(x^2-x+1)(x^2-x+1)^3((x^2-x+1)^9)^2\\\\&\equiv (x^2-x+1)(x^6-x^3+1)(x^{18}-x^9+1)^2\\\\&\equiv (x^2-x+1)(x^6-x^3+1)(x^{36}+x^{27}+x^9+1)\\\\&\equiv x^{44}-x^{43}+x^{42}-x^{41}+x^{40}-x^{39}+x^{38}-x^{37}+x^{36}+x^{35}-x^{34}+x^{33}-x^{32}+x^{31}\\\\&\qquad -x^{30}+x^{29}-x^{28}+x^{27}+x^{17}-x^{16}+x^{15}-x^{14}+x^{13}-x^{12}+x^{11}-x^{10}+x^9\\\\&\qquad +x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1\end{align}$$
It follows that the zero terms in mod $3$ are $x^{18},x^{19},x^{20},x^{21},x^{22},x^{23},x^{24},x^{25},x^{26}$.
Next, using that $(a+b+c)^2\equiv a^2+b^2+c^2\pmod 2$ repeatedly, we have, in mod $2$,
$$\begin{align}&(x^2-x+1)^{22}\\\\&=(x^2-x+1)^{16}(x^2-x+1)^4(x^2-x+1)^2\\\\&\equiv (x^{32}+x^{16}+1)(x^{8}+x^4+1)(x^4+x^2+1)\\\\&\equiv x^{44}+x^{42}+x^{38}+x^{34}+x^{32}+x^{28}+x^{26}+x^{22}+x^{18}+x^{16}+x^{12}+x^{10}+x^6+x^2+1\end{align}$$
It follows that the zero terms in mod $6$ are $x^{19},x^{20},x^{21},x^{23},x^{24},x^{25}$. 
Therefore, the total number of coefficients that are divisible by $6$ is $\color{red}{6}$.
