## Answer :

**Answer**

Given the quadratic polynomial (3x^2 - 8x + 4), we can find the sum of the roots (\(\alpha\) and \(\beta\)), the sum of their squares (\(\alpha^2 + \beta^2\)), the sum of their cubes (\(\alpha^3 + \beta^3\)), and \(3(\alpha + \beta)^2\).

1) Sum of the squares of the roots ((alpha^2 + beta^2)):

Using Vieta's formulas, the sum of the squares of the roots is \((\alpha + \beta)^2 - 2\alpha\beta\).

So, \((\alpha^2 + \beta^2) = \left(\frac{8}{3}\right)^2 - 2\cdot3\cdot4 = \frac{64}{9} - 24 = \frac{64}{9} - \frac{216}{9} = -\frac{152}{9}\).

2) Sum of the cubes of the roots (\(\alpha^3 + \beta^3\)):

Using Vieta's formulas, the sum of the cubes of the roots is \(3\alpha\beta(\alpha + \beta) - 3(\alpha + \beta)\).

So, \(\alpha^3 + \beta^3 = 3\cdot3\cdot4\cdot\frac{8}{3} - 3\cdot\frac{8}{3} = 96 - 8 = 88\).

3) \(3(\alpha + \beta)^2\):

We already found \(\alpha + \beta = \frac{8}{3}\). So, \(3(\alpha + \beta)^2 = 3\left(\frac{8}{3}\right)^2 = 3\cdot\frac{64}{9} = \frac{64}{3}\).

Therefore:

1) \(\alpha^2 + \beta^2 = -\frac{152}{9}\)

2) \(\alpha^3 + \beta^3 = 88\)

3) \(3(\alpha + \beta)^2 = \frac{64}{3}\)