Finding $\det(I+A^{100})$ where $A\in M_3(R)$ and eigenvalues of $A$ are $\{-1,0,1\}$

The eigenvalues of $A$ are $\{-1, 0, 1 \}$,

Therefore, the eigenvalues of $A^{100}$ are $\{(-1)^{100}, 0^{100}, 1^{100} \} = \{1, 0, 1 \}$ (counting multiplicities).

Therefore, the eigenvalues of $I + A^{100}$ are $\{2,1,2 \}$ (counting multiplicities).

The determinant is the product of the eigenvalues, so: $$ \det(I + A^{100}) = 1 \cdot 2^2 = 4 $$

Since $A$ has three different eigenvalues, we know it must be diagonalizable, so there is an invertible $P$ such that $$ PAP^{-1} = \begin{pmatrix}-1\\ & 0 \\ && 1 \end{pmatrix} $$

Now the determinant of any matrix such as for example $I+A^{100}$ is the same as any matrix that is is similar to, so $$ \det(I+A^{100}) = \det(P(I+A^{100})P^{-1}) = \det(PIP^{-1} + PA^{100}P^{-1}) = \det(I+ (PAP^{-1})^{100}) $$ which you can easily calculate directly.

Note that you don't actually need to know exactly what $P$ is.