Real matrix satisfying $A^3=4I_n-3A$ 
Let $A\in M_n(\mathbb{R}) $ so that $A^3=4I_n-3A$. Prove that $\det(A+I_n) =2^n$.   

My work : $A$ 's eigenvalues are the roots of $x^3+3x-4=0$, so one of the eigenvalues is $1$ and the others are $\lambda_1$ and $\lambda_2$, the roots of $x^2+x+4=0$.
Hence, 
$$\det(A+I_n) =2(\lambda_1+1)(\lambda_2+1)=2(\lambda_1  \lambda_2 +\lambda_1+\lambda_2+1) =2(4-1+1)=8$$    What is my mistake? 

Careful, you only know that the spectrum of $A$ is contained in the set of zeroes of $x^3+3x-4$.
Since $A$ is a real matrix, its minimal polynomial has real coefficients and divides $x^3+3x-4$, so it is one of 
$$x-1, x^2+x+4, (x-1)(x^2+x+4)$$
Hence the characteristic polynomial is of the form $\chi_A(x) = (x-1)^{n-2r}(x^2+x+4)^{r}$ for some $0 \le r \le \frac{n}2$.
Therefore if we denote $\lambda_1, \lambda_2$ the zeroes of $x^2+x+4$, we have
$$\det(A+I_n) = (1+1)^{n-2r}(\lambda_1+1)^r(\lambda_2+1)^r = 2^{n-2r}(\lambda_1\lambda_2+\lambda_1+\lambda_2+1)^r = 2^{n-2r}4^r = 2^n$$
