\(\left(x + \frac{1}{x} + 1\right)^2 - \left(x - \frac{1}{x} - 1\right)^2,\) We can use the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b),\) where \(a = x + \frac{1}{x} + 1 \quad \text{and} \quad b = x - \frac{1}{x} - 1.\) Calculating \( a - b \): \(a - b = \left(x + \frac{1}{x} + 1\right) - \left(x - \frac{1}{x} - 1\right) = x + \frac{1}{x} + 1 - x + \frac{1}{x} + 1 = 2 \cdot \frac{1}{x} + 2 = 2\left(\frac{1}{x} + 1\right).\) Calculating \( a + b \): \(a + b = \left(x + \frac{1}{x} + 1\right) + \left(x - \frac{1}{x} - 1\right) = x + \frac{1}{x} + 1 + x - \frac{1}{x} - 1 = 2x.\) Substituting back into the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b) = \left(2\left(\frac{1}{x} + 1\right)\right)(2x) = 4x\left(\frac{1}{x} + 1\right).\) Distributing \(4x\): \(= 4x \cdot \frac{1}{x} + 4x \cdot 1 = 4 + 4x.\) Thus, the evaluated expression is: 4 + 4x = 4(1 + x)