b. \( (A+B)^{-2}=\frac{1}{A^{2}}+\frac{1}{B^{2}} \)

Verify the identity by following steps: - step0: Verify: \(\left(A+B\right)^{-2}=\frac{1}{A^{2}}+\frac{1}{B^{2}}\) - step1: Choose a side to work on: \(\frac{1}{A^{2}+2AB+B^{2}}=\frac{1}{A^{2}}+\frac{1}{B^{2}}\) - step2: Choose the other side to work on: \(\frac{1}{A^{2}+2AB+B^{2}}=\frac{B^{2}+A^{2}}{A^{2}B^{2}}\) - step3: Verify the identity: \(\textrm{false}\) The equation \( (A+B)^{-2} = \frac{1}{A^{2}} + \frac{1}{B^{2}} \) is false. To understand why, let's analyze both sides of the equation: 1. The left side, \( (A+B)^{-2} \), represents the reciprocal of the square of the sum of \( A \) and \( B \). 2. The right side, \( \frac{1}{A^{2}} + \frac{1}{B^{2}} \), is the sum of the reciprocals of the squares of \( A \) and \( B \). These two expressions are not equivalent in general. For example, if we take specific values for \( A \) and \( B \), we can see that they yield different results. Thus, the original statement is not true.