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Consider the cubic equation

$$ x^3 + bx^2 + cx + d = 0$$

If one replaces $x$ with $x - {b\over 3}$ then this reduces to

$$ x^3 + px + q = 0$$

If one knows what to substitute with then checking the claim is easy. But my question is:

How does one see which substitution to make? How do I know that it simplifies if I replace $x$ with $x - {b\over 3}$?

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  • $\begingroup$ maybe sharply staring at pascal's triangle? $\endgroup$ – tired Sep 1 '16 at 13:37
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Recall that $-b$ is the sum of the roots of $ x^3 + bx^2 + cx + d = 0.$ That is, if the roots are $r_{1},\,r_{2},\,r_{3},$ then $(r_1 + r_2 + r_3) = -b.$ See Vieta's formulas if this is new to you.

Now note that if the graph is shifted right by $\frac{b}{3},$ which can be done by replacing $x$ with $x - \frac{b}{3},$ then each of the roots of the new equation is additively increased by $\frac{b}{3}.$ Thus, the roots of the equation you get by replacing $x$ with $x - \frac{b}{3}$ are $r_{1} + \frac{b}{3} ,\,r_{2} + \frac{b}{3},\,r_{3} + \frac{b}{3},$ and the sum of these roots is

$$\left(r_{1} + \frac{b}{3}\right) \; + \; \left(r_{2} + \frac{b}{3}\right) \; + \; \left(r_{3} + \frac{b}{3}\right) \;\; = \;\; (r_1 + r_2 + r_3) \; + \; \left(\frac{b}{3} + \frac{b}{3} + \frac{b}{3}\right) = \; -b + b = 0 $$

Since the sum of the roots of the new equation is $0,$ it follows (again, by Vieta's formulas) that the coefficient of $x^2$ in the new cubic polynomial is $0.$

In general, if you replace $x$ with $x - \frac{b}{n}$ in the equation $x^n + bx^{n-1} + \cdots = 0,$ then the coefficient of the $x^{n-1}$ term is $0$ for the polynomial you get after the replacement. For what it's worth, note also that what you're doing is shifting right by the arithmetic mean of the roots.

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  • $\begingroup$ That's a beautiful explanation / answer. Thank you very much! $\endgroup$ – student Sep 1 '16 at 19:18
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Given your third order equation $$x^3+bx^2+cx+d=0$$

let's see what a translation $x\mapsto x+x_0$ can do to it.

One has $$(x+x_0)^3+b(x+x_0)^2+c(x+x_0)+d=0$$ and so $$x^3+3x^2x_0+3xx_0^2+x_0^3+bx^2+2bxx_0+bx_0^2+cx+cx_0+d=0$$

or

$$x^3+x^2(3x_0+b)+x(3x_0^2+2bx_0+c)+(x_0^3+bx_0^2+cx_0+d)=0.$$ Now you can see what a translation can do to a polynomial of third order: it can set to zero one of the coefficients of your original equation. This is a general result and it works for each polynomial of arbitrary degrees. You can see that in order to not have the second order term you can set $$x_0=-\frac{b}{3},$$ while to get rid of the linear term you can choose $x_0$ such that $$3x_0^2+2bx_0+c=0.$$ Note that the constant term is the same initial equation for $x_0$ in place of $x$ and then setting to zero that term corresponds to solving the initial equation.

Tschirnhaus transformation on Wikipedia

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