The only polynomial in that list that doesn't split into linear factors is the one with $x^2 + x + 1$ because its discriminant is $-3$. For the other ones,
\begin{gather} \begin{aligned} x^8 & = x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\ (x-2)^3 (x^2-1) & = (x-2)(x-2)(x-2)(x-1)(x+1) \\ (x^2-1)^3 &= (x-1)(x+1)(x-1)(x+1)(x-1)(x+1) \\ \end{aligned} \end{gather} and so on. By the way, the answer $(x-3)(x^2+x+1)$ cannot be factored over the reals, but it can be factored over $\mathbb C$, because the complex numbers have this property that every polynomial splits. As an example, $ x^2 + x + 1 = \left( x - \frac{-1 + \sqrt 3 i}2 \right) \left( x - \frac{-1 - \sqrt 3 i}2 \right), $ where $i$ is defined so that $i^2 = -1$.
Hope that helps,