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Cubic Equation

Revision as of 16:11, 9 December 2020 by Vincentwant (talk | contribs)

A cubic equation is an equation of the form:

$ax^3 + bx^2 + cx + d = 0$.

A cubic equation has 3 roots, either all real OR one real, two complex.

Solving Cubic Equations

If you're too lazy to follow, look at subsection "TLDR" for each section.

Converting to a Depressed Equation

You start with the equation $ax^3 + bx^2 + cx + d = 0$.

Divide both sides by a: $x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a}$.

Now we change the coefficient of $x^2$ to $0$ (e.g. change it to a depressed cubic). We do this by substituting $y = x + \frac{b}{3a}$ or $y - \frac{b}{3a} = x$, giving:

$\left(y - \frac{b}{3a}\right)^3 + \frac{b}{a}\left(y - \frac{b}{3a}\right)^2 + \frac{c}{a}\left(y - \frac{b}{3a}\right) + \frac{d}{a} = 0$

$y^3 - \left(\frac{b}{a}\right)y^2 + \left(\frac{b^2}{3a^2}\right)y - \left(\frac{b^3}{27a^3}\right) + \left(\frac{b}{a}\right)y^2 - \left(\frac{2b^2}{3a^2}\right)y + \left(\frac{b^3}{9a^3}\right) + \left(\frac{c}{a}\right)y - \left(\frac{9abc}{27a^3}\right) + \left(\frac{27a^2d}{27a^3}\right)$

$y^3 + \left(\frac{3ac - b^2}{3a^2}\right)y + \left(\frac{2b^3 - 9abc + 27a^2d}{27a^3}\right) = 0$.

$\left(\frac{3ac - b^2}{3a^2}\right)$ is $p$ and $\left(\frac{2b^3 - 9abc + 27a^2d}{27a^3}\right)$ is $q$, so now we have $y^3 + py + q = 0$.

TLDR?

The equation is $y^3 + py + q = 0$ where $p = \left(\frac{3ac - b^2}{3a^2}\right)$ and $q = \left(\frac{2b^3 - 9abc + 27a^2d}{27a^3}\right)$.

Solving the Depressed Equation

Now here comes the smart part. Substitute $y = \sqrt[3]{u} - \sqrt[3]{v}$.

The equation becomes $u - 3\sqrt[3]{u^2v} + 3\sqrt[3]{uv^2} - v + p\sqrt[3]{u} - p\sqrt[3]{v} + q = 0$. Simplification:

$u - v + 3\sqrt[3]{uv^2} - 3\sqrt[3]{u^2v} + p\sqrt[3]{u} - p\sqrt[3]{v} + q = 0$

$u - v - (3\sqrt[3]{uv} - p)(\sqrt[3]{u} - \sqrt[3]{v}) = -q$

We want that last term to equal $0$, so we can set $3\sqrt[3]{uv} - p = 0$. (We can't use $\sqrt[3]{u} - \sqrt[3]{v} = 0$, because then $y = 0$, which is not necessarily true.) Solving this equation gives us $uv = \frac{p^3}{27}$. If $uv = \frac{p^3}{27}$, then $v - u = q$. We now have a system of equations:

$\begin{cases} uv = \frac{p^3}{27} \\ v - u = q \end{cases}$.

We can solve this via the quadratic formula. After $u$ and $v$ are obtained, we have $y = \sqrt[3]{u} - \sqrt[3]{v}$ and $x = \sqrt[3]{u} - \sqrt[3]{v} - \frac{b}{3a}$. (Note: "$\sqrt[3]{ }$" means any cube root.

TLDR?

$x = \sqrt[3]{u} - \sqrt[3]{v} - \frac{b}{3a}$ where u and v are roots of the system $\begin{cases} uv = \frac{p^3}{27} \\ v - u = q \end{cases}$.

The Cubic Formula

The cubic formula can be obtained by using the above method. These are the steps:

The depressed cubic is of the form $y^3 + \left(\frac{3ac - b^2}{3a^2}\right)y + \left(\frac{2b^3 - 9abc + 27a^2d}{27a^3}\right) = 0$.

$u$ and $v$ are the roots of the system of equations $\begin{cases} uv = \frac{p^3}{27} \\ v - u = q \end{cases}$. We can solve this by substitution:

$v = q + u$ (We are still using p and q because they might get a little messy if we use p and q in terms of a, b, c, and d.)

$u(q + u) = \frac{p^3}{27}$

$u^2 + qu - \frac{p^3}{27} = 0$

$u = \frac{-q \pm \sqrt{\frac{27q^2 - 4p^3q}{27}}}{2}$

$v = \frac{q \pm \sqrt{\frac{27q^2 - 4p^3q}{27}}}{2}$ (comes from $q + u$)

$y = \sqrt[3]{\frac{-q \pm \sqrt{\frac{27q^2 - 4p^3q}{27}}}{2}} - \sqrt[3]{\frac{q \pm \sqrt{\frac{27q^2 - 4p^3q}{27}}}{2}}$

$x = \sqrt[3]{\frac{-q \pm \sqrt{\frac{27q^2 - 4p^3q}{27}}}{2}} - \sqrt[3]{\frac{q \pm \sqrt{\frac{27q^2 - 4p^3q}{27}}}{2}} - \frac{b}{3a}$

$x = \sqrt[3]{\frac{-\left(\frac{2b^3 - 9abc + 27a^2d}{27a^3}\right) \pm \sqrt{\frac{3\left(\frac{2b^3 - 9abc + 27a^2d}{9a^3}\right)^2 - 4\left(\frac{3ac - b^2}{3a^2}\right)^3\left(\frac{2b^3 - 9abc + 27a^2d}{27a^3}\right)}{27}}}{2}}$

$- \sqrt[3]{\frac{\left(\frac{2b^3 - 9abc + 27a^2d}{27a^3}\right) \pm \sqrt{\frac{3\left(\frac{2b^3 - 9abc + 27a^2d}{9a^3}\right)^2 - 4\left(\frac{3ac - b^2}{3a^2}\right)^3\left(\frac{2b^3 - 9abc + 27a^2d}{27a^3}\right)}{27}}}{2}} - \frac{b}{3a}$

(See? I told you it would be messy.)

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