If the roots of the equation \( 3 x^{2}-5 x+1=0 \) are \( \alpha, \beta \). Find the values of; i. \( \alpha \beta^{2}+\beta \alpha^{2} \) ii. \( \alpha^{2}-\alpha \beta+\beta^{2} \) (Leaving your answer correct to 4 decimal places where appropriate).

To find the values of \( \alpha \beta^{2}+\beta \alpha^{2} \) and \( \alpha^{2}-\alpha \beta+\beta^{2} \), let’s first determine the roots \( \alpha \) and \( \beta \) of the equation \( 3x^{2}-5x+1=0 \). Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = -5, c = 1 \): 1. Calculate the discriminant: \[ D = (-5)^2 - 4 \cdot 3 \cdot 1 = 25 - 12 = 13 \] 2. Now, applying the quadratic formula: \[ x = \frac{5 \pm \sqrt{13}}{6} \] Thus, the roots are: \[ \alpha = \frac{5 + \sqrt{13}}{6}, \quad \beta = \frac{5 - \sqrt{13}}{6} \] Now we can compute \( \alpha \beta^{2} + \beta \alpha^{2} \): \[ \alpha \beta^{2} + \beta \alpha^{2} = \alpha \beta (\alpha + \beta) \] From Vieta's formulas: - \( \alpha + \beta = \frac{5}{3} \) - \( \alpha \beta = \frac{1}{3} \) Thus, \[ \alpha \beta^{2} + \beta \alpha^{2} = \alpha \beta \cdot (\alpha + \beta) = \frac{1}{3} \cdot \frac{5}{3} = \frac{5}{9} \approx 0.5556 \] Now, for \( \alpha^{2} - \alpha \beta + \beta^{2} \): This can be rewritten using the identity: \[ \alpha^{2} + \beta^{2} = (\alpha + \beta)^2 - 2\alpha\beta \] Using the values from Vieta's formulas: \[ \alpha^{2} + \beta^{2} = \left(\frac{5}{3}\right)^2 - 2 \cdot \frac{1}{3} = \frac{25}{9} - \frac{2}{3} = \frac{25}{9} - \frac{6}{9} = \frac{19}{9} \] Now, substituting back: \[ \alpha^{2} - \alpha \beta + \beta^{2} = \alpha^{2} + \beta^{2} - \alpha \beta = \frac{19}{9} - \frac{1}{3} = \frac{19}{9} - \frac{3}{9} = \frac{16}{9} \approx 1.7778 \] Thus, the final answers are: i. \( \alpha \beta^{2} + \beta \alpha^{2} \approx 0.5556 \) ii. \( \alpha^{2} - \alpha \beta + \beta^{2} \approx 1.7778 \)