8. (10 points) In the expression $(x+y+z)^{2026}+(x-y-z)^{2026}$, 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: 1028196

Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2026}+(x-t)^{2026}$. We expand both brackets using the binomial theorem and get

$$
\begin{aligned}
& (x+t)^{2026}=x^{2026}+a_{1} x^{2025} t+\ldots+a_{2025} x t^{2025}+t^{2026} \\
& (x-t)^{2026}=x^{2026}-a_{1} x^{2025} t+\ldots-a_{2025} x t^{2025}+t^{2026}
\end{aligned}
$$

Adding them up, we get

$$
(x+t)^{2026}+(x-t)^{2026}=2\left(x^{2026}+a_{2} x^{2024} t^{2}+\ldots+a_{2024} x^{2} t^{2024}+t^{2026}\right)
$$

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

$$
1+3+\ldots+2025+2027=1014^{2}=1028196
$$

## Solve the following problems with justification of the answer