Write the 2nd degree equation which have the following roots

Question

$y_1$=${(x_1+x_2\varepsilon+x_3\varepsilon^2)}^3$

$y_2$=${(x_1+x_2\varepsilon^2+x_3\varepsilon)}^3$

where $x_1,x_2,x_3$ roots for the $x^3+ax^2+bx+c=0$ and $\varepsilon$ = $\frac{-1}{2}$+$i$$\frac{\sqrt{3}}{2}$. ${\varepsilon}^3=1$

Answer 1

Hint: The coefficients of the required equation are $${(x_1+x_2\varepsilon+x_3\varepsilon^2)}^3{(x_1+x_2\varepsilon^2+x_3\varepsilon)}^3$$ and $$-{(x_1+x_2\varepsilon+x_3\varepsilon^2)}^3-{(x_1+x_2\varepsilon^2+x_3\varepsilon)}^3.$$

They are symmetric functions of $x_1,x_2,x_3$. So it remains to express them by elementary symmetric polinomials $x_1+x_2+x_3=-a$, $x_1x_2+x_2x_3+x_1x_3=b$ and $x_1x_2x_3=-c$.

Addendum: It is known that the roots of a cubic equation are of the form: $$x_1=u+v, \ \ x_2=u\varepsilon+v\varepsilon^2, \ \ x_3=u\varepsilon^2+v\varepsilon, $$ where $u,v$ are from http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method .

Then $$x_1+x_2\varepsilon+x_3\varepsilon^2=u(1+\varepsilon+\varepsilon^2)+3v=3v$$ and similarly for $x_1+x_2\varepsilon^2+x_3\varepsilon$.

$2$nd Addendum: To calculate $u^3$ and $v^ 3$ one has to reduce the equation $x^3+ax^2+bx+c=0$ to the form $y^3+py+q=0$ by the substitution $x=y-\frac{b}{3a}$. Then $$ p=-\frac{a^2}{3}+b, \ \ \ q=\frac{2a^3}{27}-\frac{ab}{3}+c, $$ $$u^3=-\frac{q}{2}+\sqrt{D}, \ \ v^3=-\frac{q}{2}-\sqrt{D}.$$

Hence $u^3+v^3=-q, \ u^3v^3=\frac{q^2}{4}-D=-\frac{p^3}{27}$ and the required equation is $$y^2+27qy-p^3=0.$$