$$(1-x+x^2)^{10}=a_0+a_1\cdot x +a_2\cdot x^2 + a_3\cdot x^3 + \cdots + a_{20}\cdot x^{20}$$
$$\implies a_0 + a_2 +a_4 +a_6 + \cdots+ a_{18} + a_{20}=?$$
I know how to expand binomials of $(a+b)^n$ type, but this one has three terms.
I can think of $(1-x+x^2)^{10}$ as $\bigr((1-x)+x^2 \bigl)^{10}$ in order to make it look like $(a+b)^n$. But I think, it would almost be impossible to find each coefficient.
How can I solve this problem?
On
Recall that like $\,e^{ix} = \cos(x) + i \sin{x},\,$ any polynomial or power series can be very simply decomposed as a sum of its even and odd parts as follows
$$f(x) = f_0(x) + f_1(x),\ \ \ {\rm where}\ \ \ \overbrace{f_0(-x) = f_0(x)}^{\large \rm even},\ \ \overbrace{f_1(-x) = -f_1(x)}^{\large \rm odd}$$
$$f_0(x) = \frac{1}2 (f(x)+f(-x))= \color{#c00}{a_0 + a_2} x^{\large 2} \color{#c00}{+ a_4} x^{\large 4} \color{#c00}{+ \cdots}$$
$$f_1(x) = \frac{1}2 (f(x)-f(-x))= a_1 + a_3 x^{\large 3} + a_5 x^{\large 5} + \cdots$$
We seek $\,\color{#c00}{a_0 + a_2 + a_4 +\cdots} = f_0(1),\ $ which for $f(x)=(1-x+x^2)^{\large 10}\,$ is
$$\begin{align} f_0(1) &= \frac{1}2(f(1)+f(-1))\\[.3em] &= \frac{1}2((1\!-\!1\!+\!1^2)^{\large 10} + (1\!-\!(-1)\!+\!(-1)^2)^{\large 10}\\[.3em] &= \frac{1}2(1^{\large 10}\!+3^{\large 10}) \end{align}\qquad\ $$
Remark $ $ As explained in the linked post, such bisections of series generalize to multisections. We can use a primitive $n$'th root of unity to construct the $m$'th $n$-section = linear progression of $\,m\!+\!kn\,$ indexed terms in an analgous manner as above.
Hint:
As we are interested in the even terms only
set $x=1,-1$ in the given identity and add to find the sum to be $$\dfrac{1+3^{10}}2$$