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.
$ay^3+(c-\frac{b^2}{3a})y+(d+\frac{2b^3}{27a^2}-\frac{bc}{2a})$
 A: 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
$$ay^3  +\left(c-\frac{b^2}{3a} \right) y +\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$$
A: I would first consider
$$\begin{align*} 0 &= a(y-t)^3 + b(y-t)^2 + c(y-t) + d \\ &= a(y^3 - 3ty^2 + 3t^2 y - t^3) + b(y^2 - 2yt + t^2) + c(y-t) + d\\ &= ay^3 + (-3at + b)y^2 + (3at^2 - 2bt + c)y + (-at^3 + bt^2 - ct + d). \end{align*}$$
Now, the value of $t$ for which the coefficient of $y^2$ equals zero obviously satisfies $3at-b = 0$, or $t = \frac{b}{3a}$.  This is not what you wrote, which is why you are experiencing..."problems."  With the correct substitution, the remaining coefficients become
$$\begin{align*}0 &= ay^3 + \bigl(\tfrac{b^2}{3a} - \tfrac{2b^2}{3a} + c\bigr)y + \bigl(-\tfrac{b^3}{27a^2} + \tfrac{b^3}{9a^2} - \tfrac{bc}{3a} + d \bigr) \\ &= ay^3 + \bigl(c - \tfrac{b^2}{3a}\bigr)y + \bigl(\tfrac{2b^3}{27a^2} - \tfrac{bc}{3a} + d\bigr).   \end{align*}$$
A: Try $$x = y - \dfrac{b}{3a}{}$$
