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

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

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

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

Now, <math>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}}}</math> (The two <math>\pm</math>'s need not be the same). As a bonus, if you make the <math>\pm</math>'s <math>+</math> and <math>-</math>, the resulting root is always real.

Substituting back, we get the cubic formula (remember <math>\frac{b}{3a}</math>): <math>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}</math>. Simplifying, we get: <math>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}</math>.

==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 <math>+</math> and <math>-</math>, 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 <math>x^3-15x-126</math>. You'll get the right answer of 6, but you need to go through a lot of stuff first. Finally, apply it on <math>x^3-6x-40</math>. What did you get? (The answer is 4). You got <math>\sqrt[3]{20-\sqrt{392}}+\sqrt[3]{20+\sqrt{392}}</math>, didn't you? Now how is that equal to 4? It is.