8. (10 points) In the expression $(x+y+z)^{2030}+(x-y-z)^{2030}$, 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: 1032256

Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2030}+(x-t)^{2030}$. We expand both brackets using the binomial theorem and get

$$
\begin{aligned}
& (x+t)^{2030}=x^{2030}+a_{1} x^{2029} t+\ldots+a_{2029} x t^{2029}+t^{2030} \\
& (x-t)^{2030}=x^{2030}-a_{1} x^{2029} t+\ldots-a_{2029} x t^{2029}+t^{2030}
\end{aligned}
$$

Adding these, we get

$$
(x+t)^{2030}+(x-t)^{2030}=2\left(x^{2030}+a_{2} x^{2028} t^{2}+\ldots+a_{2028} x^{2} t^{2028}+t^{2030}\right)
$$

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

$$
1+3+\ldots+2029+2031=1016^{2}=1032256
$$

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