I am trying to figure out the following cubic root thing.


I set up


Then I plug in for x


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.


  • $\begingroup$ If your getting bogged down in the algebra, I'd suggest first trying to perform the simplification for the special case where $a=1$. This gets rid of some of the clutter that you're struggling to digest, and the general cubic problem for $a\neq 1$ reduces to a problem of this type anyway. $\endgroup$ – David H May 16 '14 at 2:19

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

$$\begin{align*}0 &= ay^3 + \bigl(\tfrac{b^2}{3a} - \tfrac{2b^2}{3a} + c\bigr)y + \bigl(-\tfrac{b^3}{27a^2} + \tfrac{b^3}{9a^2} - \tfrac{bc}{3a} + d \bigr) \\ &= ay^3 + \bigl(c - \tfrac{b^2}{3a}\bigr)y + \bigl(\tfrac{2b^3}{27a^2} - \tfrac{bc}{3a} + d\bigr). \end{align*}$$

  • $\begingroup$ yes that makes sense I see I initially wanted to write t=b/3a but I typed it in latex wrong $\endgroup$ – Fernando Martinez May 16 '14 at 2:45
  • $\begingroup$ I see you have $(-3at+b)y^2$ on the part before the substitution is (-1) factored out I am wondering? $\endgroup$ – Fernando Martinez May 16 '14 at 3:17

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 $$ay^3 +\left(c-\frac{b^2}{3a} \right) y +\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$$


Try $$x = y - \dfrac{b}{3a}{}$$

  • $\begingroup$ Note that even with your formula you substituted incorrectly. I assume this was just a typo. $\endgroup$ – Brad May 16 '14 at 2:43
  • $\begingroup$ The OP has already used the substitution $x = y - \frac{b}{3a}$, so what exactly is the point in repeating this and posing it as a hint? $\endgroup$ – beep-boop Jul 2 '14 at 22:22

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