<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] Method $\#1:$
Following your process, $$(1+x+x^2)^{10}=\left(\frac{1-x^3}{1-x}\right)^{10} =(1-x^3)^{10}(1-x)^{-10}$$
$$=\left[1-\binom{10}1x^3+\binom{10}2x^6-\cdots\right]\left[1-10(-x)+\frac{-10(-10-1)}2(-x)^2+\cdots\right]$$
Clearly the coefficient of $x^2$ is $\displaystyle \frac{-10(-10-1)}2=55$
Method $\#2:$
For a clearer version of user$125053$'s answer
$$(1+x+x^2)^{10}=\left[(1+x)+x^2\right]^{10}$$
$$=\binom{10}0(1+x)^{10}+\binom{10}1(1+x)^9(x^2)^1+\binom{10}1(1+x)^8(x^2)^2\cdots$$
So, the required coefficient of $x^2$ is
the coefficient of $x^2$ in $\displaystyle\binom{10}0(1+x)^{10}+$ the coefficient of $x^0$ in $\displaystyle\binom{10}1(1+x)^9$
$\displaystyle=\binom{10}0\cdot\binom{10}2+\binom{10}1\cdot\binom90 =55 $ again
Method $\#3:$
$$(1+x+x^2)^{10}=\left[1+x(1+x)\right]^{10}$$
$$=1+\binom{10}1x(1+x)+\binom{10}2\{x(1+x)\}^2+\binom{10}3\{x(1+x)\}^3+\cdots$$
$$=1+\binom{10}1(x+x^2)+\binom{10}2\{x^2(1+x)^2\}+\binom{10}3\{x(1+x)\}^3+\cdots$$
So, the required coefficient of $x^2$ is  $\displaystyle\binom{10}1+\binom{10}2=55$
Method $\#4:$
Using Multinomial Theorem,
the general term of the expansion of $\displaystyle(1+x+x^2)^{10}$
is $\displaystyle\frac{10!}{a!b!c!}(1)^a(x)^b(x^2)^c=\frac{10!}{a!b!c!}x^{b+2c}$  where $a,b,c\ge0$ and $a+b+c=10$
Now we need $b+2c=2\iff2c=2-b\le2$ as $b\ge0$
$\displaystyle\implies b\le1\implies b=0,1$
If $b=0,\cdots$
If $b=1,\cdots$
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