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Let \(\frac{x + a}{x - a} = y\) ∴ Substituting \(y\) into the given equation, we get: \[ y^2 - 5y + 6 = 0 \] Or, \[ y^2 - (3 + 2)y + 6 = 0 \Rightarrow y^2 - 3y - 2y + 6 = 0 \Rightarrow y(y - 3) - 2(y - 3) = 0 \Rightarrow (y - 3)(y - 2) = 0 \] So, either \(y = 3\) or \(y = 2\) When \(y = 3\), \[ \frac{x + a}{x - a} = 3 \Rightarrow 3x - 3a = x + a \Rightarrow 2x = 4a \Rightarrow x = 2a \] When \(y = 2\), \[ \frac{x + a}{x - a} = 2 \Rightarrow 2x - 2a = x + a \Rightarrow x = 3a \] ∴ The required solutions are \(x = 2a\) and \(x = 3a\)