Let's carefully prove that: \[ (1 - e^2)(1 + e^2) = 1 \] --- ## Step 1: Recognize the structure The expression resembles a difference of squares: \[ (a - b)(a + b) = a^2 - b^2 \] In our case: - \(a = 1\) - \(b = e^2\) --- ## Step 2: Apply the difference of squares formula Using the formula: \[ (1 - e^2)(1 + e^2) = 1^2 - (e^2)^2 \] Calculating each term: \[ 1^2 = 1 \] \[ (e^2)^2 = e^{2 \times 2} = e^4 \] Thus: \[ (1 - e^2)(1 + e^2) = 1 - e^4 \] --- ## Step 3: Show that the expression equals 1 The original statement claims: \[ (1 - e^2)(1 + e^2) = 1 \] From the calculation, the left side simplifies to: \[ 1 - e^4 \] For this to be equal to 1: \[ 1 - e^4 = 1 \] Subtract 1 from both sides: \[ - e^4 = \] \[ e^4 = \] --- ## **Final conclusion:** - The equality \((1 - e^2)(1 + e^2) = 1\) **only holds** if \(e^4 = \), which implies: \[ e = \] - For **any other value** of \(e\), the statement **does not hold**. --- ## **Answer:** \[ \boxed{ \text{The equality }(1 - e^2)(1 + e^2) = 1 \text{ holds only when } e=. } \]