Cubic formula

The cubic formula is a very complicated formula used to solve cubics. It is not used very often, as schools don't teach it and problem writers usually hide a simpler tactic instead.

Cardano's formula

Start with the cubic $ax^3+bx^2+cs+d$ . The constant $a$ controls how steep it is. The constant $d$ shifts it up and down. The constant $c$ controls the slope of the middle. And the constant $b$ shifts it left to right. For now, let's just reduce it to $x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}$ , as everyone knows how to do that. Consider the cubic $(x+\frac{b}{3a})^3=x^3+\frac{b}{a}x^2+\frac{b^2}{3a^2}x+\frac{b^3}{27a^3}$ . The first two terms match, and if we substitute $x$ with $x-\frac{b}{3a}$ , we get rid of that annoying $x^2$ term. We'll just add it back at the end. Making the substitution results in: $\left(x-\frac{b}{3a}\right)^3+\frac{b}{a}\left(x-\frac{b}{3a}\right)^2+\frac{c}{a}\left(x-\frac{b}{3a}\right)+\frac{d}{a}=x^3+\frac{3ac-b^2}{3a^2}x+\frac{2b^3+27a^2d-9abc}{27a^3}$ .

By now, let's define $p$ and $q$ as $p=\frac{3ac-b^2}{3a^2}$ and $q=\frac{2b^3+27a^2d-9abc}{27a^3}$ . We get the depressed cubic $x^3+px+q=0$ or $x^3=-px-q$ . First, let's expand $(u+v)^3=u^3+3u^2v+3uv^2+v^3=3uv(u+v)+(u^3+v^3)$ . Now, let $x=u+v$ so $(u+v)^3$ matches up with $x^3$ , $3uv(u+v)$ matches up with $-px$ , and $(v^3+w^3)$ matches up with $-q$ . Things are about to get exciting!

Now, $x=u+v$ , $uv=-\frac{p}{3}$ , and $u^3+v^3=-q$ . Let's cube p to get $u^3v^3=-\frac{p^3}{27}$ and then write the equation $v^3\cdot -q=v^3(v^3+u^3)=v^6+v^3u^3=v^6-\frac{p^3}{27}$ . This may not sound exciting until our substitution $w=v^3$ . We got rid of $u$ and now we have $w^2+\left(q\right)w+\left(-\frac{p^3}{27}\right)$

Using the quadratic formula, $w=\frac{-q}{2}+\pm\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$ , so $v=\sqrt[3]{\frac{-q}{2}+\pm\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$ and by symmetry, so is $u$ . (If this results in $\frac{0}{0}$ , interpret it as $0$ ).

Now, $x=\sqrt[3]{\frac{-q}{2}+\pm\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{\frac{-q}{2}+\pm\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$ (The two $\pm$ 's need not be the same). As a bonus, if you make the $\pm$ 's $+$ and $-$ , the resulting root is always real.

Substituting back, we get the cubic formula (remember $\frac{b}{3a}$ ): $x=\sqrt[3]{\frac{\frac{-2b^3-27a^2d+9abc}{27a^3}}{2}\pm\sqrt{\frac{\left(\frac{2b^3+27a^2d-9abc}{27a^3}\right)^2}{4}+\frac{\left(\frac{3ac-b^2}{3a^2}\right)^3}{27}}}+\sqrt[3]{\frac{\frac{-2b^3-27a^2d+9abc}{27a^3}}{2}\pm\sqrt{\frac{\left(\frac{2b^3+27a^2d-9abc}{27a^3}\right)^2}{4}+\frac{\left(\frac{3ac-b^2}{3a^2}\right)^3}{27}}}-\frac{b}{3a}$ . Simplifying, we get: $x=\sqrt[3]{\frac{-2b^3-27a^2d+9abc}{54a^3}\pm\sqrt{\left(\frac{2b^3+27a^2d-9abc}{54a^3}\right)^2+\left(\frac{3ac-b^2}{9a^2}\right)^3}}\\+\sqrt[3]{\frac{-2b^3-27a^2d+9abc}{54a^3}\pm\sqrt{\left(\frac{2b^3+27a^2d-9abc}{54a^3}\right)^2+\left(\frac{3ac-b^2}{9a^2}\right)^3}}\\-\frac{b}{3a}$ .

Disadvantages

First of all, even with normal cubics, the square rooting step introduces complex numbers, which can make the cube root more complicated. In the end, if you take $+$ and $-$ , they will be conjugates but it will go through a awful mess first. It is possible to make another cubic formula that avoids this insanity, but that requires trig. Next, try applying it on $x^3-15x-126$ . You'll get the right answer of 6, but you need to go through a lot of stuff first. Finally, apply it on $x^3-6x-40$ . What did you get? (The answer is 4). You got $\sqrt[3]{20-\sqrt{392}}+\sqrt[3]{20+\sqrt{392}}$ , didn't you? Now how is that equal to 4? It is.

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Cubic formula

Cardano's formula

Disadvantages