Solving cubic equation? I am trying to figure out the following cubic root thing.
$ax^3+bx^2+cx+d=0$
I set up 
$x=y-\frac{3}{ba}$
Then I plug in for x
$a(y-\frac{3}{ba})^3+b(y-\frac{3}{3a})^2+c(y-\frac{3}{ba})=0$
The issue I am having trouble with is the simplification 
I try to multiply it all but I gets messy.
Maybe the binomial theorem can be used
this is supposed to go down into the depressed cubic which is.
$ay^3+(c-\frac{b^2}{3a})y+(d+\frac{2b^3}{27a^2}-\frac{bc}{2a})$

I think you may have the substitution wrong. Try
$$x=y-\frac{b}{3a}$$
then 
$$a\left(y-\frac{b}{3a}\right)^3 +b\left(y-\frac{b}{3a}\right)^2 +c\left(y-\frac{b}{3a}\right)+d=0$$ gives
$$ay^3 -by^2 +\frac{b^2}{3a}y -\frac{b^3}{27a^2}+by^2-\frac{2b^2}{3a}y +\frac{b^3}{9a^2}+cy-\frac{bc}{3a}+d=0$$
or
$$ay^3  +\left(c-\frac{b^2}{3a} \right) y +\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$$
