the geometric explain of $t = x-\frac{a}{3}$ in the simplify of cubic equation $x^3+ax^2+bx+c=0$ Assume $$f(x) = x^3+ax^2+bx+c$$
we have $$f''(x)=2a+6x$$. we get $x = -\frac{a}{3}$
Magically, If we take the transformation: $$t = x -\left(-\frac{a}{3}\right)$$. we can transform the above equation into $$f(x)=g(t) = t^3+\left(b-\frac{a^2}{3}\right)t+\frac{2a}{27}-\frac{ab}{3}+c$$
Why ? why this transform cancellation the quadratic term? Can anyone give an Geometric explain?  Thanks very much. 

Let's look at the quadratic term of $f(t - \dfrac{a}{3})$. It came from $2$ sources. One is the $2$nd term of the cubic expansion $(t - \dfrac{a}{3})^3$: $-3t^2\cdot \dfrac{a}{3}$ which is: $-at^2$, and the other term comes from the $1$st term of $a(t - \dfrac{a}{3})^2$ which is: $at^2$. So the coefficient of $t^2$ is: $-a + a = 0$. So there is no quadratic term $t^2$.
