8. (10 points) In the expression $(x+y+z)^{2028}+(x-y-z)^{2028}$, 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: 1030225

Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2028}+(x-t)^{2028}$. We expand both brackets using the binomial theorem and get

$$
\begin{aligned}
& (x+t)^{2028}=x^{2028}+a_{1} x^{2027} t+\ldots+a_{2027} x t^{2027}+t^{2028} \\
& (x-t)^{2028}=x^{2028}-a_{1} x^{2027} t+\ldots-a_{2027} x t^{2027}+t^{2028}
\end{aligned}
$$

Adding them up, we get

$$
(x+t)^{2028}+(x-t)^{2028}=2\left(x^{2028}+a_{2} x^{2026} t^{2}+\ldots+a_{2026} x^{2} t^{2026}+t^{2028}\right)
$$

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

$$
1+3+\ldots+2027+2029=1015^{2}=1030225
$$

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