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?
 A: 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. 
A: If $a,b,c$ are the eigenvalues of $A$, then $\det(A)=abc$, and $\det(A+dI)=(a+d)(b+d)(c+d)$.
So we are given that
$$
(a+1)(b+1)(c+1)=\det (A+I)=\det (A+2I)=(a+2)(b+2)(c+2),
$$
and hence
$$
(ab+bc+ca)+(a+b+c)+1=2(ab+bc+ca)+4(a+b+c)+8,
$$
or simpler
$$
(ab+bc+ca)+3(a+b+c)+7=0.\tag{1}
$$
We need to show that
$$
2\det (A+I)+\det (A-I)+6=3\det(A)
$$
which is equivalent to
$$
2(abc+(ab+bc+ca)+(a+b+c)+1)+(abc-(ab+bc+ca)+(a+b+c)-1)+6=3abc,
$$
or simpler
$$
(ab+bc+ca)+3(a+b+c)+7=0\tag{2}
$$
Clearly $(1)$ and $(2)$ are identical and hence we are done.
