If $A$ is square matrix such that ${A^2} = A$, then show that ${(I + A)^3} = 7A + I$.

Since, ${A^2} = A$ and $(I + A) \cdot (I + A) = {I^2} + IA + AI + {A^2}$

$= {I^2} + 2AI + {A^2}$

$= I + 2A + A = I + 3A$

and $(I + A) \cdot (I + A)(I + A) = (I + A)(I + 3A)$

$= {I^2} + 3AI + AI + 3{A^2}$
$= I + 4AI + 3A$

$= I + 7A = 7A + I$