For $(1 + x + x^2)^6$, find the term which has $x^6$ in it. 
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.

I tried to use Newton's binomial formula as:
$$
(1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right)
$$
and that's all I can think of, other then just to compute it.

Apply the trinomial theorem: $$(1+x+x^2)^6=\sum_{a+b+c=6,\:0\le a,b,c\le 6} \binom{6}{a,b,c}(1)^a(x)^b(x^2)^c$$
To get the $x^6$ term, the possibilities are $(a,b,c)=(3,0,3),(2,2,2),(1,4,1),(0,6,0)$. Hence the $x^6$ term is $$\binom{6}{3,0,3}x^6+\binom{6}{2,2,2}x^6+\binom{6}{1,4,1}x^6+\binom{6}{0,6,0}x^6=141x^6$$
