1. Compute the coefficient of $x^{80}$ in the polynomial $\left(1+x+x^{2}+\cdots+x^{80}\right)^{3}$ after combining like terms.
Solution: $\left(1+x+x^{2}+\cdots+x^{80}\right)^{3}=\left(1+x+x^{2}+\cdots+x^{80}\right)\left(1+x+x^{2}+\cdots+x^{80}\right)\left(1+x+x^{2}+\cdots+x^{80}\right)$

Let after expanding the brackets, the term $x^{p} \cdot x^{q} \cdot x^{r}$ is obtained if the factor $x^{p}$ is chosen from the first bracket, $x^{q}$ - from the second bracket, $x^{r}$ - from the third bracket. Then $x^{p} \cdot x^{q} \cdot x^{r}=x^{80}, p+q+r=80$, $0 \leq p, q, r \leq 80, p, q, r \in N$. We need to calculate the number of such terms after expanding the brackets. Let $p=0$, the number of different sums $q+r=80$ is 81. Let $p=1$, the number of different sums $q+r=79$ is 80. And so on. Let $p=80$, the number of different sums $q+r=0$ is 1. Calculate the total number of such terms: $1+2+3+\cdots+81=\frac{82 \cdot 81}{2}=3321$.

Answer: 3321.