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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 $$\tag{1} x^3 + ax^2 +bx+c=0.$$

Putting $y=x + \frac{a}{3}$, we can translate $(1)$ into $$\tag{2} y^3 +3hy + k =0.$$ Next, if we write $y=u+v$, then $u+v$ is a root of the cubic $$\tag{3} y^3 -3uvy - (u^3 + v^3).$$

Equating coefficients in $(2)$ and $(3)$ gives the following formula for finding the roots of $(2)$: $$\tag{4} \sqrt[3]{\frac{1}{2}\left(-k + \sqrt{k^2 + 4h^3}\right)}+ \sqrt[3]{\frac{1}{2}\left(-k - \sqrt{k^2 + 4h^3}\right)} .$$

Now, if $u$ is one of the cube roots of $${\frac{1}{2}\left(-k + \sqrt{k^2 + 4h^3}\right)} $$ then the other two roots are $u\omega$ and $u{\omega}^2$ (where $\omega=e^{\frac{2i\pi}{3}}$) and, since $v=-\frac{h}{u}$, the three roots of $(2)$ are $$ \tag{5} u-\frac{h}{u},\quad u\omega - \frac{h\omega^2}{u}, \quad u\omega^2-\frac{h\omega}{u}.$$

Question: I know I am missing something painfully obvious, but how does $v=-\frac{h}{u}$ give the last two roots in $(5)$?

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  • $\begingroup$ Compare (2) and (3) to see $3h= - 3uv$. $\endgroup$ – GEdgar Aug 17 '18 at 0:23
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There's an error there: $\omega=\exp\left(\frac{2\pi i}3\right)$, not $\exp\left(\frac{\pi i}3\right)$. Therefore, $\omega^2=\frac1\omega$ and $\omega=\frac1{\omega^2}$. So,$$-\frac h{\omega u}=-\frac{\omega^2}u\text{ and }-\frac1{\omega^2u}=-\frac\omega u.$$

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  • $\begingroup$ Aah, of course. Thanks! $\endgroup$ – Moed Pol Bollo Aug 17 '18 at 0:19

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