If A is a square matrix and I is the identity matrix of same order such that $A^2 = I$, then $(A - I)^3 + (A + I)^3 - 3A$ is equal to

Given that $A$ is a square matrix, $I$ is the identity matrix of the same order, and $A^2 = I$.

To find: $(A - I)^3 + (A + I)^3 - 3A$

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Expanding $(A - I)^3$:

$(A - I)^3 = A^3 - 3A^2I + 3AI^2 - I^3$

Since $AI = IA = A$, $I^2 = I$, and $I^3 = I$:

$(A - I)^3 = A^3 - 3A^2 + 3A - I$

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Expanding $(A + I)^3$:

$(A + I)^3 = A^3 + 3A^2I + 3AI^2 + I^3$

$(A + I)^3 = A^3 + 3A^2 + 3A + I$

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Adding both expansions:

$(A - I)^3 + (A + I)^3 = (A^3 - 3A^2 + 3A - I) + (A^3 + 3A^2 + 3A + I)$

$= 2A^3 + 6A$

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Substituting into the original expression:

$(A - I)^3 + (A + I)^3 - 3A = 2A^3 + 6A - 3A$

$= 2A^3 + 3A$

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Using the condition $A^2 = I$:

$A^3 = A \times A^2$

$= A \times I$

$= A$

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Therefore:

$2A^3 + 3A = 2A + 3A$

$= 5A$

The answer is $5A$.