Is my simple (in my opinion) way of solving cubic equations correct?

I've been analyzing ways of solving cubic equations and I've come up with this one.

I've tried to make it as simple as possible.

So I'll show you a way of solving cubic equations when none of the roots are rational. When at least some of them are, you can simply use the Rational Root test.

When I have an equation $ax^3 + bx^2 + cx + d = 0$, instead of $x$ I put $y - {b \over 3a}$.

Then I get this equation: $y^3 + py + q = 0$.

I've found that then $y$ equals to:

$$y_1 = {{ \sqrt[3]{12 \sqrt{12p^3+81q^2}-108q} - \sqrt[3] {12 \sqrt{12p^3 + 81q^2 }+108q} } \over 6}$$

And the other solutions are got when we divide $ax^3 + bx^2 + cx + d$ by $(y - y_1)$ and solve a quadratic equation.

Then subtract $b \over 3a$ from every $y$ to get all 3 solutions $x$.

This way is great in a way that it's short and simple to understand. You need to know or have the formula written down somewhere, though. I've memorized it.

• A starter on writing $\LaTeX$ on the site is here x^2 works fine when put between dollar signs: $x^2$ For multi-character superscripts you put them in braces, so x^{123} gives $x^{123}$ To get cube roots, you use \sqrt[3]{a+b} to get $\sqrt[3]{a+b}$ – Ross Millikan Nov 10 '13 at 1:32
• How do I give 'useful comment' votes? I'm completely new here, sorry. – user26486 Nov 10 '13 at 2:54
• @mathh Hover your mouse over the comment and a triangle will appear to the left of the comment, above a flag. Click the triangle to upvote the comment. – Alexander Gruber Nov 10 '13 at 3:43
• I can't see any flags or triangles. – user26486 Nov 10 '13 at 4:05
• That's probably because my current reputation is really low. – user26486 Nov 10 '13 at 4:18

The substitution you're doing is called a Tschirnhaus transformation. You're right in that it always gives you something of the form $$y^3+py+q=0,$$ and in fact, we can compute formulae for $p$ and $q$ by $$p=\frac{3ac-b^2}{3a^2}\hspace{30pt}q=\frac{2b^3-9abc+27a^2d}{27a^3}.$$ Unfortunately, the formula you give for $y_1$ doesn't always work. Take for example $$f(x)=x^3+x+1.$$ In this case, $p=q=1$ (since $b=0$, the substitution is trivial). However, your formula for $y_1$ produces $$y_1=\frac{1}{6} \left(\sqrt[3]{12 \sqrt{93}+108}-\sqrt[3]{12 \sqrt{93}-108}\right)$$ Plugging this into $f$ gives you $f(y_1)=2$, so your method doesn't work here. However, I notice that this $y_1$ is the real root of $x^3+x-1$, which may help you locate the error in your formula.
EDIT Now that the formula has been updated, take $$f(x)=x^3+4x^2+4x+1.$$ Here we get $p=-4/3$, $q=11/27$, so $12 p^3+81 q^2=-15<0$, so you're gonna have to cube root something with an imaginary part. What do you do when this happens?
My prediction is that this method will only work when this quantity turns out to be non-negative, i.e. $4 a c^3 + 4 b^3 d + 27 a^2 d^2 > b c (b c + 18 a d)$.
• @mathh $q=1$ here, so the change doesn't affect the calculation in this case, right? – Alexander Gruber Nov 10 '13 at 3:47