Writing Cubic Equation in terms of discriminant (with possible shifts and translations) So I noticed this fact for the following fact for quadratic equations. I need one notation that if one equation can be gotten from another through a shift or scaling of variable then I will denote them as equivalent, using the symbol $\leftrightarrow$ :
\begin{align}( ax^{2}+bx+c&\leftrightarrow 4a[( ax^{2}+bx+c )] = ( 4a^{2}x^{2}+4abx+4ac ) = [(2ax)^{2} + 2b(2ax) + 4ac]\\&\leftrightarrow x^{2}+2bx+4ac\\&  \leftrightarrow [(x-b)^{2} + 2b(x-b) + 4ac ] = x^{2} - (b^{2}-4ac) \end{align}
This means we have managed to show $( ax^{2}+bx+c ) \leftrightarrow x^{2} - r$ , where r is discriminant. 
I was wondering if we could do something similar for depressed cubic ( $ x^{3} + px + q) $ and relate its discriminant ( $ -4p^{3} - 27q^{2} $ ) through the same process of shifting and scaling of the variable 'x'. 
 A: No, we cannot. If a cubic equation is equivalent to an equation of the form $x^3-r$, then either it has only one root (if $r=0$) or it has $3$ distinct roots (if $r\neq0$). But then consider the equation $x^3-3x+2$, which has $2$ roots: $1$ and $-2$.
A: Problem 1: Shifting (except by 0) and scaling (except by 1) do not preserve the property of being a depressed cubic with leading coefficient 1.  The change of variable $x \mapsto ax+b$ changes $x^3+px+q$ to $a^3x^3+3a^2bx^2 + \cdots$. If you want this to be a depressed cubic with leading coefficient 1 (such as $x^3 - c$) then you need $a^3 = 1$ and $3a^2b = 0$, so $a=1$ and $b=0$, which is not a change of variables at all. 
Problem 2: Shifting and scaling using rational coefficients keeps a real root real and a nonreal root nonreal. The polynomial $x^3 - c$ for nonzero rational $c$ has one real root and two distinct nonreal roots in $\mathbf C$. The polynomial $x^3 - 9x - 9$ has three distinct real roots. Therefore you can’t turn $x^3 - 9x - 9$ into $x^3 - c$ by shifting and scaling with rational coefficients. 
There are higher-level problems, related to Galois theory, but the above two problems already show why cubics can’t be treated in exactly the same way as quadratics. Once your cubic polynomial is a depressed cubic, you have done as much as you can by tricks that come from the quadratic case.
