<s>[INST] 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.
 [/INST] 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 $x_2=\frac{1}{3}$
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