Given $A \in \mathbb{R}^{3 \times 3}$, prove the next equality:
$$A^3-\text{tr}(A)A^2+\text{tr}(\text{adj}(A))A-\det(A)I\_3=0.$$

We know that:

$$A^3-\text{tr}(A)A^2+\text{tr}(\text{adj}(A))A-\det(A)I\_3=\\A^3-\sum\_{i=1}^n a\_{ii}A^2+\sum\_{i=1}^n b\_{ii}A-\det(A)I\_3\implies\\A^3-\sum\_{i=1}^n a\_{ii}A^2+\sum\_{i=1}^n b\_{ii}A=\det(A)I\_3\implies \\A^3-\left(\sum\_{i=1}^n a\_{ii}A+\sum\_{i=1}^n b\_{ii}\right)A=\det(A)I\_3.$$

Here I'm too stuck. Can someone help me?