If $A^3+2A^2+2A+I_n=0_n$, then $\det(A)=-1$, where $A$ is a square matrix of odd size, with real entries, and $0_n$ is the zero matrix. Let us denote the identity matrix by $I_n$.
What I have managed to obtain so far:
The case $A=-I_n$ is straightforward.
Obviously, the case $A \neq -I_n$ is troublesome. So far, by writing $A(A^2+2A+2I_n)=-I_n, $ I have obtained that $A$ is nonsingular.
Also $(A+I_n)(A^2+A+I_n)=0_n, $ from which $\det(A+I_n)=0=\det(A-\epsilon I_n)=\det(A-\bar{\epsilon}I_n)=\det(A^2+A+I_n), $ where ${\epsilon}^2+\epsilon + 1 =0 .$ This is because, by Sylvester's Inequality for ranks, we have $\operatorname{rank}(A^2+A+I_n) \leq n-1,$ from $(A+I_n)(A^2+A+I_n)=0_n$ and $\operatorname{rank}(A+I_n)\geq 1$. Because $p_A$ has real coefficients, both $det(A-\bar{\epsilon}I_n)=0$ and $\det(A-{\epsilon}I_n)=0$ always take place.
This means that the characteristic polynomial of $A$, $p_A \in \mathbb R[X], $ has $p_A(\epsilon)=p_A(\bar{\epsilon})=p_A(-1)=0. $ To determine the determinant of $A$, I should be able to compute $p_A(0).$ I have tried writing $p_A(0)$ as a linear combination of $p_A(\epsilon),p_A(\bar{\epsilon}),p_A(-1)$ however to no avail.
Since each eigenvalue of $A$ solves $(\lambda+1)(\lambda^2+\lambda+1)=0$, the possible eigenvalues are $-1,\,\exp\frac{\pm2\pi i}{3}$. Since $\det A\in\Bbb R$, each non-real eigenvalue is used equally many times. But these eigenvalues' product is $1$, so they have no effect on $\det A$. Now an odd number of copies of $-1$ are used, making the final result $-1$.