If A is a square matrix such that ${A^2} = I$, then ${(A - I)^3} + {(A + I)^3} - 7A$ is equal to

We have, ${A^2} = I$
$\therefore$ ${(A - I)^3} + {(A + I)^3} - 7A = \left[ {(A - I) + (A + I)\left\{ {{{(A - I)}^2}} \right.} \right.$

$\left. {\left. { + {{(A + I)}^2} - (A - I)(A + I)} \right\}} \right] - 7A$

$= \left[ {(2A)\left\{ {{A^2} + {I^2} - 2AI + {A^2} + {I^2} + AI - \left( {{A^2} - {I^2}} \right)} \right\}} \right] - 7A$

$= 2A\left[ {I + {I^2} + I + {I^2} - {A^2} + {I^2}} \right] - 7A$

$= 2A[5I - I] - 7A$
$= 8AI - 7AI$
$= AI = A$