8. (10 points) In the expression $(x+y+z)^{2020}+(x-y-z)^{2020}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained?
Answer: 1022121

Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2020}+(x-t)^{2020}$. We expand both brackets using the binomial theorem and get

$$
\begin{aligned}
& (x+t)^{2020}=x^{2020}+a_{1} x^{2019} t+\ldots+a_{2019} x t^{2019}+t^{2020} \\
& (x-t)^{2020}=x^{2020}-a_{1} x^{2019} t+\ldots-a_{2019} x t^{2019}+t^{2020}
\end{aligned}
$$

Adding these, we get

$$
(x+t)^{2020}+(x-t)^{2020}=2\left(x^{2020}+a_{2} x^{2018} t^{2}+\ldots+a_{2018} x^{2} t^{2018}+t^{2020}\right)
$$

Notice that when expanding different $t_{1}^{n}$ and $t_{2}^{n}$, we will get different monomials (since the degrees of $x$ will be different). Also, when expanding $t^{n}$, we will get $n+1$ different monomials, so the final answer will be

$$
1+3+\ldots+2019+2021=1011^{2}=1022121
$$

## Solve the following problems with justification of the answer