<s>[INST] coefficient of $x^2$, in $(1+x+x^2)^{10}$ 
How  to find coefficient of $x^2$, in $(1+x+x^2)^{10}$, without actually expanding it?

I think the fact $\dfrac{1-x^3}{1-x}=1+x+x^2$ may help.
But can't use it!
 [/INST] As you noted:
$$
    [x^2]\left(1+x+x^2\right)^{10} = [x^2] \frac{(1-x^3)^{10}}{(1-x)^{10}}
$$
Because $x^3$ will only contribute to terms $x^3$ and higher order
$$
  [x^2] \frac{(1-x^3)^{10}}{(1-x)^{10}} = [x^2] \frac{1}{(1-x)^{10}} = [x^2] \frac{1}{(1-x)^{10}}
$$
Now using $n \cdot [x^n] g(x) = [x^{n-1}] g^\prime(x)$ with $n=2$ and $g(x) = \frac{1}{9}(1-x)^{-9}$:
$$
  [x^2] \frac{1}{(1-x)^{10}} = 3 [x^{3}] \frac{1}{9 (1-x)^9} = 3 \cdot 4 [x^4] \frac{1}{9 \cdot 8 (1-x)^8} = \frac{(11)!}{2} \cdot [x^{11}] \frac{1}{9! \cdot (1-x)} 
$$
Using $[x^n] \left(1-x\right)^{-1} = 1$ for $n\geq 1$ we arrive at:
$$
   [x^2] \frac{1}{(1-x)^{10}} = \frac{(11)!}{2 \cdot 9!} = \frac{11 \cdot 10}{2} = 55
$$
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