If A is a square matrix such that A²=A, then (I-A)³+A is equal to

Given that A is  a square matrix  such that  $A^2=A$
$\begin{array}{l}(\mathrm{I}-\mathrm{A}{)}^3+\mathrm{A}=(\mathrm{I}-\mathrm{A}{)}^2(\mathrm{I}-\mathrm{A})+\mathrm{A}\\ =\left({\mathrm{I}}^2-2\mathrm{AI}+{\mathrm{A}}^2\right)(\mathrm{I}-\mathrm{A})+\mathrm{A}\\ =(\mathrm{I}-2\mathrm{A}+\mathrm{A})(\mathrm{I}-\mathrm{A})+\mathrm{A}\left(\therefore {\mathrm{A}}^2=\mathrm{A},{\mathrm{I}}^2=\mathrm{I}\right)\\ =(\mathrm{I}-\mathrm{A})(\mathrm{I}-\mathrm{A})+\mathrm{A}\end{array}$
$\begin{array}{l}=(\mathrm{I}-\mathrm{A}{)}^2+\mathrm{A}\\ ={\mathrm{I}}^2-2\mathrm{AI}+{\mathrm{A}}^2+\mathrm{A}\\ =\mathrm{I}-2\mathrm{A}+\mathrm{A}+\mathrm{A}\left(\because {\mathrm{A}}^2=\mathrm{A},{\mathrm{I}}^2=\mathrm{I}\right)\\ (\mathrm{I}-\mathrm{A}{)}^3+\mathrm{A}=\mathrm{I}\end{array}$
Therefore, Option (A) I is correct option