Is there a property similar to $ x^2 = (x-1)(x+1)+1$ for $ x^3$? I am looking for a way to decompose $x^3$ in a similar way. 
 A: As others have pointed out:
$$x^3 = (x-1)(x^2+x+1)+1$$
You say you want three factors.  You can do this but you must use non–real complex numbers.  You can write
$$x^n = (x -\alpha_1)(x-\alpha_2)...(x-\alpha_n) + 1$$
where the $\alpha_i$ are the $n^\mathrm{th}$ roots of unity.  
A: I think so: $x^3 = (x-1)(x^2+x+1) + 1$
A: We can write $$x^3= (x-1)(x^2+x+1)+1$$  Also, another representation can be $$x^3 = (x+1)(x^2-x+1)-1$$
A: $$\small \begin{vmatrix} x+n & 1 & 0 & 0 & \ddots & 0 & 0 & 0 & 0\\ -n & x+n-2 & 2 & 0 & \ddots & 0 & 0 & 0 & 0\\ 0 & -n+1 & x+n-4 & 3 & \ddots & 0 & 0 & 0 & 0\\ 0 & 0 & -n+2 & x+n-6 & \ddots & 0 & 0 & 0 & 0\\ \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots\\ 0 & 0 & 0 & 0 & \ddots & x-n+6 & n-2 & 0 & 0\\ 0 & 0 & 0 & 0 & \ddots & -3 & x-n+4 & n-1 & 0\\ 0 & 0 & 0 & 0 & \ddots & 0 & -2 & x-n+2 & n\\ 0 & 0 & 0 & 0 & \ddots & 0 & 0 & -1 & x-n \end{vmatrix} = x^{n+1}$$
It goes without saying that $n$ is obviously an integer, since the determinant is always polynomial in $x$. By setting $n=1$, your decomposition can be recovered.
