An equation for the third powers of the roots of a given quadradic polynomial The roots of the equation $3x^2-4x+1=0$ are $\alpha$ and $\beta$. Find the equation with integer coefficients that has roots $\alpha^3$ and $\beta^3$.
GIVEN
SR: $\alpha + \beta = \frac43$
PR: $\alpha\beta = \frac13$
REQUIRED
SR: $\alpha^3 + \beta^3 = (\alpha + \beta) (\alpha^2 + 2\alpha \beta + \beta^2)$
$\alpha^3 + 3\alpha^2 \beta + \alpha \beta^2 + \alpha^2 \beta + 3\alpha \beta^2 + \beta^3$
$\alpha^3 + 3\alpha^2 \beta + \alpha \beta^2 + \alpha^2 \beta + 3\alpha \beta^2 + \beta^3 = (\alpha + \beta)^3$
$\alpha^3+ \beta^3 = (\alpha + \beta)^3 - 3(\alpha \beta)^2 - (\alpha \beta)^2$
$\alpha^3 + \beta^3 = (\frac43)^2 - 3(\frac13)^2 - (\frac13)^2$
$\alpha^3 + \beta^3 = \frac{16}{9} - \frac13 - \frac19$
$\alpha^3 + \beta^3 = \frac43$
PR: $\alpha^3 \beta^3$
$(\alpha \beta)^3 = (\frac13)^3 = \frac{1}{27}$
EQUATION
$x^2 - \frac43 x + \frac{1}{27} = 0$
$27x^2 - 36x + 1 = 0$
I am not sure if it's suppose to be:
$(\alpha + \beta) (\alpha^2 + 2\alpha \beta + \beta^2)$
Or
$(\alpha + \beta) (\alpha^2 + 3\alpha \beta + \beta^2)$
Also sorry about the formatting, if you don't mind fixing it for me. Not sure how to do it.
Thank You.

Using the quadratic formula, we have $$\alpha, \beta = \frac{-(-4) \pm \sqrt{(-4)^2-4(3)(1)}}{6} \\ = \frac{4 \pm \sqrt{16-12}}{6} \\ = \frac{4 \pm 2}{6} \\ = 1, \frac{1}{3} $$ Hence a polynomial you could use with roots at $\alpha^3 = 1^3 = 1$ and $\beta^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$ would be $$f(x) = (x-1)\left(x-\frac{1}{27}\right) \\ = x^2 -\frac{28x}{27}+\frac{1}{27}$$ Or in general, for any constant $ C \neq 0$, then $C\cdot f(x)$ will have zeroes at $\alpha^3, \beta^3$
