jmc

We could check to see which divisors of $-35$ are roots of the cubic $x^3-9x^2+27x-35=0$.

However, notice that $x^3-9x^2+27x-35=(x-3{)}^3-2^3$. As such, we can factor this as a difference of cubes: $(x-3{)}^3-2^3=((x-3)-2)((x-3{)}^2+2(x-3)+2^2)=(x-5)(x^2-4x+7)$.

We see that $x^2-4x+7$ cannot be factored any further, so our answer is $\boxed{(x-5)(x^2-4x+7)}$.