Prove statement about determinants. $A$ is a $3\times 3$ matrix over $\mathbb{R}$, I want to show that if $$\det(A + I_3)=\det(A+2I_3),$$ then $$2\det(A+I_3) + \det(A-I_3) + 6 = 3\det A.$$ 
Can you help me?

First show that you can assume $A$ diagonal. Then the question reduces to proving a basic identity between two polynomials. 
More explicitely, you can write $\det (A+ x I_3) = \sigma_3 + x \sigma_2 + x^2 \sigma_1 + x^3$, where $\sigma_3=a_1 a_2 a_3$, $\sigma_2 = a_1 a_2 + a_1 a_3 + a_2 a_3$ and $\sigma_1 = a_1 + a_2 + a_3$, if the $a_i$ are the diagonal elements of $A$. The assumption gives you $\sigma_2 = -3 \sigma_1 - 7$, and the wanted equality follows readily. 
