If $A$ is a square matrix such that $A^2=I$, then find the simplified value of $(A-I{)}^3+(A+I{)}^3-7A$.

Apply the formula $a^3+b^3={\left(a+b\right)}^3-3ab\left(a+b\right)$ in ${\left(A-I\right)}^3+{\left(A+I\right)}^3-7A$.

$={\left(\left(A-I\right)+\left(A+I\right)\right)}^3-3\left(A-I\right)\left(A+I\right)\left(\left(A-I\right)+\left(A+I\right)\right)-7A$

$={\left(2A\right)}^3-3\left(A^2-I\right)\left(2A\right)-7A$

$=8A^3-6A^3+6A-7A$

$=2A^3–A$

$=2A^2\cdot A–A$

$=2A–A$

$=A$