\(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\) = (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3) x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0 = x(x + 3)9x + 4) > 0 Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0 = x > -4 (solution only) Case 2 (+, -, -) = x > 0, x + 4 0, x -3, x 0 = x -4 = x < -0, -4 < x < -3(solution) combining the solutions -4 < x < -3