Let \(\cfrac{x+4}{x-4} = y\) ∴ Substituting \(\cfrac{x+4}{x-4}\) with \(y\) in the given equation, we get: \(y^2 - 5y + 6 = 0\) i.e., \(y^2 - (3 + 2)y + 6 = 0\) i.e., \(y^2 - 3y - 2y + 6 = 0\) i.e., \(y(y - 3) - 2(y - 3) = 0\) i.e., \((y - 3)(y - 2) = 0\) So, either \(y - 3 = 0 ⇒ y = 3\) or \(y - 2 = 0 ⇒ y = 2\) When \(y = 3\), then \(\cfrac{x+4}{x-4} = 3\) i.e., \(3x - 12 = x + 4\) i.e., \(3x - x = 4 + 12\) i.e., \(2x = 16\) i.e., \(x = 8\) Again, when \(y = 2\), then \(\cfrac{x+4}{x-4} = 2\) i.e., \(2x - 8 = x + 4\) i.e., \(2x - x = 4 + 8\) i.e., \(x = 12\) \(\therefore\) The required solutions are \(x = 8\) and \(x = 12\) (Answer)