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Difference between revisions of "Cubic formula"

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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.
 
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.
 +
 +
[[category:Algebra

Revision as of 12:17, 27 September 2024

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 $aw^3+bw^2+cw+d$. 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 let $w$ be $x-\frac{b}{3a}$, 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 because they become complex conjugates.

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.

[[category:Algebra

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