If A and B are two equal ordered matrix such that AB = BA

then prove that

1. (A+B) (A-B) = A²-B²
2. (A-B)²= A²-2AB+B²
3. (A+B)³= A³+3A²B+3AB²+B³

given : AB = BA
​1.
LHS of the given equation is:
$$\begin{gathered} (A+B)(A-B)=A^2+BA-AB-B^2 \\ =A^2+AB-AB-B^2 \\ =A^2-B^2 \end{gathered}$$
= RHS
2.
LHS of the given equation is:
$$\begin{gathered} (A-B{)}^2=(A-B\left)\right(A-B) \\ =A^2-BA-AB+B^2 \\ =A^2-AB-AB+B^2 \\ =A^2-2AB+B^2 \end{gathered}$$
= RHS
3.
LHS of the given equation is:
$$\begin{gathered} (A+B{)}^3=(A+B\left)\right(A+B{)}^2 \\ =(A+B)(A^2+2AB+B^2) [\sin ce AB=BA] \\ =A^3+2A^{2}B+ABB+BAA+2BAB+B^3 \\ =A^3+2A^{2}B+A{B}^2+ABA+2ABB+B^3 [\sin ce AB = BA] \\ =A^3+2A^{2}B+A{B}^2+AAB+2A{B}^2+B^3 \\ =A^3+2A^{2}B+3A{B}^2+A^{2}B+B^3 \\ =A^3+3A^{2}B+3A{B}^2+B^3 \end{gathered}$$
= RHS
hope this helps you.