Vtyjkyuk

I’m going through the derivation for the Cubic Formula.

Quickly passing through the steps so that there is no cunfusion in notation:

We consider the cubic equation
$$tag1 x^3 + ax^2 +bx+c=0.$$

Putting $y=x + fraca3$, we can translate $(1)$ into $$tag2 y^3 +3hy + k =0.$$ Next, if we write $y=u+v$, then $u+v$ is a root of the cubic $$tag3 y^3 -3uvy - (u^3 + v^3).$$

Equating coefficients in $(2)$ and $(3)$ gives the following formula for finding the roots of $(2)$:
$$tag4 sqrt[3]frac12left(-k + sqrtk^2 + 4h^3right)+ sqrt[3]frac12left(-k - sqrtk^2 + 4h^3right) .$$

Now, if $u$ is one of the cube roots of $$frac12left(-k + sqrtk^2 + 4h^3right) $$ then the other two roots are $uomega$ and $uomega^2$ (where $omega=e^frac2ipi3$) and, since $v=-frachu$, the three roots of $(2)$ are $$ tag5 u-frachu,quad uomega - frachomega^2u, quad uomega^2-frachomegau.$$

Question: I know I am missing something painfully obvious, but how does $v=-frachu$ give the last two roots in $(5)$?

There's an error there: $omega=expleft(frac2pi i3right)$, not $expleft(fracpi i3right)$. Therefore, $omega^2=frac1omega$ and $omega=frac1omega^2$. So,$$-frac homega u=-fracomega^2utext and -frac1omega^2u=-fracomega u.$$