8. (10 points) In the expression $(x+y+z)^{2022}+(x-y-z)^{2022}$, 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: 1024144

Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2022}+(x-t)^{2022}$. We expand both brackets using the binomial theorem and get

$$
\begin{aligned}
& (x+t)^{2022}=x^{2022}+a_{1} x^{2021} t+\ldots+a_{2021} x t^{2021}+t^{2022} \\
& (x-t)^{2022}=x^{2022}-a_{1} x^{2021} t+\ldots-a_{2021} x t^{2021}+t^{2022}
\end{aligned}
$$

Adding them up, we get

$$
(x+t)^{2022}+(x-t)^{2022}=2\left(x^{2022}+a_{2} x^{2020} t^{2}+\ldots+a_{2020} x^{2} t^{2020}+t^{2022}\right)
$$

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

$$
1+3+\ldots+2021+2023=1012^{2}=1024144
$$

## Solve the following problems with justification of the answer