Let $A,B$ be squared matrices. Given $A=I-AB$, Prove: $B^3=0$ if and only if $A=I-B+B^2$ 
Let $A,B$ be squared matrices. 
  Given $A=I-AB$,
Prove: $B^3=0  \iff A=I-B+B^2$ 

The question has 3 sections: 

  
*
  
*Given $A=I-AB$, Prove that $A$ is invertible and that $BA=AB$. 
  
*Prove that if $B$ is a symmetric matrix, then $A$ is symmetric
  
*Prove: $A=I-B+B^2$ if and only if $B^3=0$. 

I proved the first two sections, and did the first direction of the third section, Thus we suppose that $A=I-B+B^2$, and I proved that $B^3=0$ as follows: 
Suppose $A=I-B+B^2$, then $A=I-AB=I-B+B^2$
$$-AB=-B+B^2$$
$$\text{ we'll substitute $A$ by $I-B+B^2$}$$
$$-(I-B+B^2)B= -B+B^2$$
$$-B+B^2-B^3 =-B+B^2 \Longrightarrow B^3 =0$$
As wished. 
Now let $B^3=0$. Prove: $A=I-B+B^2$. 
$$AB = I-A$$
We'll multiply by $B$ on the left side:
$$AB^2 = B-AB$$
$$AB^2+AB=B$$
$$AB(B+I)=B$$
and from this point I'm stuck. 

For the second case, just do iteration of the defintion of $A$.
$A=I-AB=I-(I-AB)B=I-B+AB^2=I-B+(I-AB)B^2$
$A=I-B+B^2-AB^3$
But $B^3=O$, then you have your definition of $A$
