How to come up with ways of simplifying equations? 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 $x - {b\over 3}$?

 A: 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
A: 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.
