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.
On
First it's good to understand how the coefficients of a polynomial relate to the roots. Suppose for now that your polynomial factorises as
$f = (x-e_1)(x-e_2)(x-e_3)$.
Multiplying out, you find
$f = x^3 - (e_1+e_2+e_3)x^2 + (e_1e_2 + e_1e_3 + e_2e_3)x - e_1e_2e_3$.
So the coefficient of the $x^2$ term of any cubic polynomial is minus the sum of its roots. (In fact, the coefficient of the $x^{n-1}$ term in any polynomial of degree $n$ is minus the sum of the roots.) In your example, the number $-a/3$ is the mean of the three roots.
You asked for a geometric interpretation of why that particular transformation makes the $x^2$ term go away. Your question can be viewed as the following: how can I shift the graph $y=f(x)$ horizontally to make the sum (or, equivalently, the mean) of the roots zero? Well, shifting the graph, say, left by $7$ shifts all three roots left by $7$ and also shifts their mean left by $7$. So to achieve what you want, you find the mean of the three roots, and shift so that the mean gets taken to zero. That's what the transformation $x \mapsto x - (-a/3)$ does for you.
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 $t^2$.