FRACTURE

For some reason, the cubic formula keeps pulling my attention. I know the final expression is famous for being complicated, but that is exactly what makes me curious.

Formulas do not appear out of thin air. Somewhere behind the intimidating symbols there must be a sequence of simple ideas, each leading naturally to the next.

Before attempting the monster cubic formula, I want to become comfortable with the cube roots of unity. They seem to be one of the hidden structures that keep appearing whenever cubics are involved.

Every cubic seems complicated until it reveals its factors.

Let, $x^3 = 1 \implies x^3 = 1^3 \implies x^3 - 1^3 = 0 \implies (x-1)(x^2+x+1) = 0$

Factoring immediately exposes the real root and the pair of complex roots hiding behind the quadratic factor.

$$ \begin{aligned} &x-1 = 0 \qquad \qquad &x^2+x+1 = 0 \\ &\therefore \ \boxed{x = 1} \qquad \qquad &\therefore \ \boxed{x = -\frac{1}{2} \pm \frac{i\sqrt{3}}{2}} \end{aligned} $$

Let, $\displaystyle \omega_1 = 1; \quad \omega_2 = -\frac{1}{2} + \frac{i\sqrt{3}}{2}; \quad \omega_3 = -\frac{1}{2} - \frac{i\sqrt{3}}{2}$

The interesting part is not that these roots exist, but how tightly they are connected to one another.

$\bullet \quad \displaystyle \omega_2^2 = \left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^2 = \frac{-3+1-2\sqrt{3}i}{4} = -\frac{1}{2} - \frac{i\sqrt{3}}{2} = \omega_3$

$\bullet \quad \displaystyle \omega_3^2 = \left(-\frac{1}{2} - \frac{i\sqrt{3}}{2}\right)^2 = \frac{-3+1+2\sqrt{3}i}{4} = -\frac{1}{2} + \frac{i\sqrt{3}}{2} = \omega_2$

$\bullet \quad \displaystyle \frac{1}{\omega_2} = \frac{1}{-\dfrac{1}{2} + \dfrac{i\sqrt{3}}{2}} = \frac{2}{i\sqrt{3}-1} = \frac{2(i\sqrt{3}+1)}{-4} = -\frac{1}{2} - \frac{i\sqrt{3}}{2} = \omega_3$

$\bullet \quad \displaystyle \frac{1}{\omega_3} = \frac{1}{-\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2}} = \frac{-2}{i\sqrt{3}+1} = \frac{-2(i\sqrt{3}-1)}{-4} = -\frac{1}{2} + \frac{i\sqrt{3}}{2} = \omega_2$

$\bullet \quad \displaystyle \omega_1\omega_2\omega_3 = 1 \times \frac{-1+i\sqrt{3}}{2} \times \frac{-1-i\sqrt{3}}{2} = \frac{1+3}{4} = 1$

$\bullet \quad \displaystyle \omega_1+\omega_2+\omega_3 = 1 + \frac{-1+i\sqrt{3}}{2} + \frac{-1-i\sqrt{3}}{2} = \frac{2-2}{2} = 0$

Squaring swaps them. Taking reciprocals swaps them. Multiplying them together returns 1. And the sum of all three roots is 0. These roots behave less like separate numbers and more like different faces of the same object.

Nothing fundamentally changes when 1 is replaced by $a$. The entire structure simply stretches by a factor of $a$.

$\displaystyle x^3 = a^3 \implies x^3 - a^3 = 0 \implies (x-a)(x^2+ax+a^2) = 0$

$\displaystyle \therefore \ x = a, \ \frac{-a}{2} \pm \frac{ai\sqrt{3}}{2} \implies x = a, \left(-\frac{1}{2} \pm \frac{i\sqrt{3}}{2}\right)a$

$$ \begin{align*} \displaystyle \therefore \ \boxed{x = a, \ \omega a, \ \omega^2 a} \qquad \qquad &\left| \ \displaystyle \omega = -\frac{1}{2} + \frac{i\sqrt{3}}{2}\right. \\ &\left| \ \displaystyle \omega^2 = -\frac{1}{2} - \frac{i\sqrt{3}}{2}\right. \end{align*} $$

The cube roots of unity are acting like a template: once they are known, the roots of $x^3=a^3$ follow immediately.

The cube roots of unity feel like the first pieces of a much larger puzzle.

Before attempting the cubic formula, I want to first derive the quadratic formula from scratch using the $y+k$ substitution. The quadratic case is simple enough that I can focus entirely on the mechanics of the substitution itself, which will reappear in a more sophisticated form when I eventually tackle the cubic monster.

The Quadratic Formula