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I have a general question about the cubic equation:

Let $a,b,c \in \mathbb{C}$ with $a \neq 0$. The general cubic equation is $t^3+at^{2}+bt+c = 0$. To get Cardano's Formula, we first transform the equation so that $a= 0$ (i.e. $t^3+bt+c = 0$). Then we reduce this to a quadratic that we can solve. What is the motivation behind the transformations that reduce $t^3+at^2+bt+c = 0$ to $t^3+bt+c = 0$?

In particular, if we let $y = t+ \frac{a}{3}$, then $t = y-\frac{a}{3}$ which cancels the $at^2$ term. This is called a Tschirnhaus transformation. We are left with an equation of the form $$y^3+py+q = 0$$

But how do we know what transformations to use to get Cardano's Formula, where $$p = \frac{a^2-2a^3+3b}{3}\quad\mathrm{and}\quad q = \frac{2a^3-9ab+27c}{27}\quad ?$$ Finally if we let $y = \sqrt[3]{u}+ \sqrt[3]{v}$ we eventually get a quadratic which leads to Cardano's Formula.

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    If you want motivation, I think Lagrange's derivation of the cubic formula is much better motivated (the motivation comes from Galois theory): http://en.wikipedia.org/wiki/Cubic_function#Lagrange.27s_method2011-07-06
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    Yes, sorry about that.2011-07-06
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    I haven't read Tschirnhaus, have Cardano. Dissecting cubes was used by him to get what we would now think of as identities for homogeneous cubics. May have been motivated by similar strategies for quadratics that go back to al-Khwarizmi. The elimination of $x^2$ was presumably motivated by the success of del Ferro in solving at least one non-trivial class of cubics with the $x^2$ term missing.2011-07-06
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    @user: +1. To be more specific, del Ferro solved the classes of equations $x^3=px+q$ and $x^3+q=px$ for positive (integers) $p$ and $q$.2011-07-07
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    @Didier Piau: Thanks, I thought the speculation was $x^3+px=q$ with uncertainty about $x^3=px+q$. Worried about this a number of years ago, couldn't locate anything but very second-hand stuff.2011-07-07
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    @Didier Piau: Didn't del Ferro solve $x^3 = px+q$, $x^3 + px= q$, and $x^3+q = px$?2011-07-07
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    @Damien What do you mean by the "motivation" for the change of variable $t = y - a/3$? Do you mean (1) why do we want to eliminate the quadratic term, or (2) how do we know that that substitution will do that?2014-06-21

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