Express the following as a product of polynomials in $a$:

$$
a^{10}+a^{5}+1
$$
The terms of our expression form a three-term geometric sequence with 1 as the first term and $a^{5}$ as the common ratio. Therefore, in the case of $a^{5} \neq 1$ (i.e., $\left.a \neq 1\right)$, using the known sum formula, and repeatedly applying the identity for the difference of equal exponents, the expression can be transformed as follows:

$$
\begin{aligned}
1+a^{5}+a^{10}= & \frac{\left(a^{5}\right)^{3}-1}{a^{5}-1}=\frac{\left(a^{3}\right)^{5}-1^{5}}{a^{5}-1^{5}}=\frac{\left(a^{3}-1\right)\left(a^{12}+a^{9}+a^{6}+a^{3}+1\right)}{(a-1)\left(a^{4}+a^{3}+a^{2}+a+1\right)}= \\
= & \frac{(a-1)\left(a^{2}+a+1\right)\left(a^{12}+a^{9}+a^{6}+a^{3}+1\right)}{(a-1)\left(a^{4}+a^{3}+a^{2}+a+1\right)}= \\
& =\left(a^{2}+a+1\right) \frac{a^{12}+a^{9}+a^{6}+a^{3}+1}{a^{4}+a^{3}+a^{2}+a+1}
\end{aligned}
$$

In the last form, performing the division in the quotient yields a polynomial, based on which

$$
a^{10}+a^{5}+1=\left(a^{2}+a+1\right)\left(a^{8}-a^{7}+a^{5}-a^{4}+a^{3}-a+1\right)
$$

and thus we have met the requirements of the problem.

Our transformation is not valid for the case $a=1$, but direct substitution shows that the equality also holds for this exceptional case.