Let $B $ be a $3\times 3$ matrix.
argue invertibalitiy o $A = B^4 + 3 B^2 + 7 B + 3 I$
I guess getting it to product form and trying on determinant will work by dont konw how to I wish someone would know.
by the way Im intereset if there is any other way
Your matrix can be written as $A = p(B)$ where $p$ is the polynomial $$ p(x) = x^4 + 3x^2 + 7x + 1 $$ By the spectral mapping theorem, the matrix $A$ will be invertible if and only if none of the eigenvalues of $B$ is a solution to $p(x) = 0$. That is, $A$ will be invertible if and only if none of these values is an eigenvalue of $B$.
Moreover, if we know all $3$ eigenvalues $\lambda_i$ of $B$, we may compute this new determinant as $$ \det(A) = p(\lambda_1)p(\lambda_2)p(\lambda_3) $$