[tex]\displaystyle\int_8^{10}(x-8)(x-9)(x-10)\,\mathrm dx[/tex]
Consider the substitution,
[tex]y=x-9\implies\begin{cases}y-1=x-10\\y+1=x-8\\\mathrm dy=\mathrm dx\end{cases}[/tex]
so that the integral is equivalent to
[tex]\displaystyle\int_{-1}^1y(y+1)(y-1)\,\mathrm dy[/tex]
Notice that
[tex]f(y)=y(y+1)(y-1)\implies f(-y)=-y(-y+1)(-y-1)=-y(y-1)(y+1)=-f(y)[/tex]
which means [tex]f(y)[/tex] is odd, so
[tex]\implies\displaystyle\int_{-1}^1f(y)\,\mathrm dy=0[/tex]
Perhaps more work than necessary, but it does make it easier to see that the original integrand exhibits some symmetry about [tex]x=9[/tex].
