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Let $A \in \mathcal{M}_n(\mathbb{R})$ such as $A^3=4\mathbb{I}_n-3A$.Prove that $$\det(A+\mathbb{I}_n)=2^n$$

$A$ satisfies the polynomial equation $x^3+3x-4=(x^2+x+4)(x-1)=0.$ The minimal polynomial is a factor of that, and the characteristic polynomial contains no factors that aren't also present in the minimal polynomial. And a real matrix has a real characteristic polynomial. So the characteristic polynomial for $A$ is $p(x)=(x^2+x+4)^k(x-1)^{n-2k}$ for some $0\le k\le \frac n2.$ We note also that $\det(A-x \mathbb{I})=(-1)^np(x).$ Hence:

\begin{align*}
\det(A+\mathbb{I})&=(-1)^np(-1)\\
&=(-1)^n(1-1+4)^k(-2)^{n-2k}\\
&=(-1)^{2n-2k}4^k2^{n-2k}\\
&=2^n
\end{align*}

that is , what we desire.