Solution—
We have
$$
\begin{aligned}
&\left(a^{6}-b^{6}\right) \\
=&\left(a^{3}\right)^{2}-\left(b^{3}\right)^{2} \\
=&\left(a^{3}-b^{3}\right)\left(a^{3}+b^{3}\right) \\
=&(a-b)\left(a^{2}+a b+b^{2}\right)(a+b)\left(a^{2}-a b+b^{2}\right) \\
=&(a-b)(a+b)\left(a^{2}+a b+b^{2}\right)\left(a^{2}-a b+b^{2}\right) . \\
\therefore \quad &\left(a^{6}-b^{6}\right)=(a-b)(a+b)\left(a^{2}+a b+b^{2}\right)\left(a^{2}-a b+b^{2}\right) .
\end{aligned}
$$
