Answer
(a)
$$(A-1)(A+1) = A^2+A-A-1 = A^2-1$$
$$(A-1)(A^2+A+1) = A^3+A^2+A-A^2-A-1= A^3-1$$
$$(A-1)(A^3+A^2+A+1)= A^4+A^3+A^2+A -A^3-A^2-A-1 =A^4-1 $$
(b)
$$A^5-1= (A-1)(A^4+A^3+A^2+A+1) = A^5+A^4+A^3+A^2+A-A^4-A^3-A^2-A-1 = A^5-1$$
$$A^n-1 = (A-1)(..(n-1)\times..+A^{n-3}+A^{n-2}+A^{n-1}+1)$$
Work Step by Step
(a) We will simply multiply right-hand side of these expressions and simplify them as possible:
$(A-1)(A+1) = A^2+A-A-1 = A^2-1$
$(A-1)(A^2+A+1) = A^3+A^2+A-A^2-A-1= A^3-1$
$(A-1)(A^3+A^2+A+1)= A^4+A^3+A^2+A -A^3-A^2-A-1 =A^4-1 $
(b) According to the sequence we got in (a), we can consider that:
$A^5-1$ will factor as $(A-1)(A^4+A^3+A^2+A+1)$ To verify, let's multiply and simplify this expression:
$=> A^5+A^4+A^3+A^2+A-A^4-A^3-A^2-A-1 = A^5-1$
Now, according to the pattern we faced, we can generalize a factoring formula:
$$A^n-1 = (A-1)(..(n-1)\times..+A^{n-3}+A^{n-2}+A^{n-1}+1)$$
