8. (10 points) In the expression $(x+y+z)^{2034}+(x-y-z)^{2034}$, 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: 1036324

Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2034}+(x-t)^{2034}$. We expand both brackets using the binomial theorem and get

$$
\begin{aligned}
& (x+t)^{2034}=x^{2034}+a_{1} x^{2033} t+\ldots+a_{2033} x t^{2033}+t^{2034} \\
& (x-t)^{2034}=x^{2034}-a_{1} x^{2033} t+\ldots-a_{2033} x t^{2033}+t^{2034}
\end{aligned}
$$

Adding these, we get

$$
(x+t)^{2034}+(x-t)^{2034}=2\left(x^{2034}+a_{2} x^{2032} t^{2}+\ldots+a_{2032} x^{2} t^{2032}+t^{2034}\right)
$$

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

$$
1+3+\ldots+2033+2035=1018^{2}=1036324
$$

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