An equation for the third powers of the roots of a given quadradic polynomial

Question

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.

Answer 1

solving the equation we get $x_1=1$ and $x_2=\frac{1}{3}$ thus we have $\alpha^3=1$ and $\beta^3=\frac{1}{27}$ and our equation is given by $(x-1)(x-1/27)=0$ therefore we get $27x^2-28x+1=0$ the equation $3x^2-4x+1=0$ is equivalent to $x^2-\frac{4}{3}x+\frac{1}{3}=0$ after the formula for a quadratic equation we obtain $x_{1,2}=\frac{2}{3}\pm\sqrt{\frac{4}{9}-\frac{3}{9}}$ thus we get $x_1=1$ or $x_2=\frac{1}{3}$

Answer 2

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 $\alpha^3, \beta^3$