# Solving the Cubic Equation (using Lagrange Resolvents)

This is from my textbook. I am having trouble working out the calculations that the author skips over.

So we start with the polynomial $\ X^3 - aX^2 + bX -c$ with zeros $x_1,x_2,x_3$. Then we define $u_1=x_1 + \omega x_2 + \omega^2 x_3$ and $u_2=x_1+\omega^2 x_2 +\omega x_3$. We show that $u_1 u_2$ and $(u_1)^3 + (u_2)^3$ are symmetric polynomials so can be expressed in terms of $a,b,c$. The next line is where I got stuck.

$$(u_1)^3 + (u_2)^3 = 2(x_1^3+x_2^3+x_3^3) - 3(x_1^2x_2 +x_1x_2^2+x_1^2x_3+x_1x_3^2+x_2^2x_3+x_2x_3^2)$$

When I multiplied out $(u_1)^3+(u_2)^3$, I got the following: $$2(x_1^3+x_2^3+x_3^3) +(3\omega + 3\omega^2)(x_1^2x_2+x_1x_2^2+x_1^2x_3+x_1x_3^2+x_2^2x_3+x_2x_3^2) + 12x_1x_2x_3$$

I'm not sure if there is a way I can simplify $3\omega + 3\omega^2$ and I'm also confused why the $x_1x_2x_3$ term goes away.

• Hi, what textbook are (were, perhaps) you reading? – zar Oct 20 '16 at 9:26

It looks to me like the $12x_1x_2x_3$ term is correct and might be a typo in your source (it's symmetric so you should be able to proceed).
Note that since $\omega$ is a primitive cube root of unity its minimal polynomial is $\omega^2+\omega+1 = 0$, so $3\omega+3\omega^2 = -3$, as desired. The identity $\omega^2+\omega+1 = 0$ also tells us that the sum of the coefficients of $u_1, u_2$ are zero, so we know that the same is true for $u_1^3+u_2^3$, further evidence for the $12x_1x_2x_3$ term being correct.