I have tried finding a solution to this problem myself and I have failed. It's just a challenge for me. Could you please tell me how far am I in solving this question?

My approach for quadratic equation :

I am still a student, and when I first saw the quadratic formula I wanted to derive it myself and I did it. This was my logic: the equation $x^2 + bx + c=0$ is obviously supposed to contain 2 roots (real or imaginary). So any 2 points in an axes can be represented as: $n+k$ or $n-k$. so (1) : putting $x = n+k$, (2): putting $x=n-k$ we get two equations; on subtracting $(1) - (2)$ we get an equation in $n$ and $k$ as, $4nk+2kb=0$ which gives $n=-b/2$. Now substituting this value of $n$ in $(1)$ and $(2)$ and adding them we get $k= (+or-) \sqrt{b^2 -4c}/2 $.

My approaches for trying to solve cubic equations:

Let $x^3+bx^2+cx+d = 0$ be the equation. My first trials were focused on extracting the roots by various substitutions. Let me explain some of them. Originally, when I derived the quadratic equation, I wanted to literally "represent the solution" by $n+k$ and $n-k$. most methods that I have thought of for cubic equation are similar to that of the quadratic (like assigning 3 unambiguous variables to uniquely represent 3 roots- $n$ and $k$ represents unique values), by considering 3 unique values, but they end up being ambiguous. Then I was suddenly struck with the idea that I could define certain random operators to "transform" the equation in some way changing the coefficients of $x^2, x, x^0$. I have defined the first transformation as: $x^3+bx^2+cx+d=0 \to x^3+ b'x^2 + c'x+d' =0 $ then $b' = b$, $c' =b^2+c$, $d' = bc-d$. Applying this operation changes the modified equation's zeros as follows: $\alpha'=\frac{\alpha+\beta}{2}, \beta' = \frac{\beta+\gamma}{2}, \gamma' = \frac{\alpha+\gamma}{2} $ if the initial zeros are $\alpha, \beta, \gamma$. My latest assumption is that creating $n$ number of transformations and applying them can help us obtain the roots in some combination.

The other operation I have defined is "increment roots by by $\theta$ operation". in this case, $x^3+ bx^2+ cx+ d \to x^3+b'x^2+ c'x+d=0$. here, $b' = b + 3\theta, c' = 3\theta^2+ 2\theta b+ c, d' = \theta^3 +b\theta^2 + c\theta + d$. This modification transforms the original roots $\alpha, \beta, \gamma$ to $\alpha' = \alpha + \theta, \beta' = \beta + \theta, \gamma' = \gamma + \theta$

Now for my questions:

How far am I from solving the cubic polynomial problem myself without any external help?

More over, I give up. How to find the roots of a cubic equation?

Could my latest idea for solving cubic equations actually be applied (the transformation method)?

Is it possible to find the roots with my transformation method? (my second transformation definition was successful in converting the coefficient b or c to zero by transforming the roots linearly)

  • 1
    $\begingroup$ If (after a while) you want much more information about cubics, please see this. $\endgroup$ Jun 22, 2014 at 13:43
  • $\begingroup$ Would you like a new answer to your question, given you only have one answer? $\endgroup$ Dec 26, 2015 at 3:35
  • 1
    $\begingroup$ Did you know there exists a formulae for the solution to the quartic equation? I note it is too large for a comment and hard to find on the internet, but an algebraic solution does exist. $\endgroup$ Dec 26, 2015 at 3:37

2 Answers 2


I'll give you some hints.

  1. Transform the equation to the form $x^3 + 3bx + c$. (The 3 just comes from a simple change of variable, but will come in handy). Your transformation method may come in handy here.

  2. Now for the clever trick. Let $x = w - \frac{b}{w}$. This transforms the equation into $w^3 + c - \frac{b^3}{w^3} = 0$, which is a quadratic in $w^3$, and can be solved.

  3. You now have a value for $w$. Can you now find a root of the equation?


Let us begin with


What we can do is substitute $x=y-\frac{b}{3a}$


Relying on the binomial theorem, we can expand the terms.

$$a(y^3-\frac{b}{a}y^2+\frac{b^2}{3a^2}y-\frac{b^3}{27a^3})+b(y^2-\frac{2b}{3a}y+\frac{b^2}{9a^2})+c(y-\frac{b}{3a})+d=ay^3-by^2+\frac{b^2}{3a}y-\frac{b^3}{27a^2}+by^2-\frac{2b^2}{3a}y+\frac{b^3}{9a^2}+cy-\frac{bc}{3a}+d=0 $$

Note that $-by^2+by^2=0$

Now, collect the terms

$$ay^3+(c+\frac{b^2}{3a}-\frac{2b^2}{3a})y+d-\frac{b^3}{27a^2}+\frac{b^3}{9a^2}-\frac{bc}{3a}=0 $$

We now have a cubic trinomial in $y$ whose quadratic coefficient is 0.

Let us set the following



$$Q=d-\frac{b^3}{27a^2}+\frac{b^3}{9a^2}-\frac{bc}{3a} $$

With substitution, we get


Divide by $a$


Now let


And substitute


Now here is something brilliant. $3u^2v+3uv^2=3uv(u+v)$



We can set $u$ and $v$ such that $3uv+\frac{P}{a}=0$ Then $u^3+v^3+\frac{Q}{a}=0$


Cube both sides


Divide by $27$.


And note that

$$u^3+v^3=-\frac{Q}{a} $$

Divide $u^3v^3=-\frac{P^3}{27a^3}$ by $u^3$

$$ v^3=-\frac{P^3}{27a^3u^3}$$

Now substitute for $v^3$


Multiply by $27a^3u^3$


Add $27a^2Qu^3$


We have a sextic equation, yes. But we also have a quadratic equation in terms of $u^3$.


Note that we could have divided $u^3v^3=-\frac{P^3}{27a^3}$ by $v^3$ instead of $u^3$and ultimately reached this same quadratic equation in terms of $v^3$ So we can set


$$v^3=\frac{-Q}{2a}-\sqrt{\frac{Q^2}{4a^2}+\frac{P^3}{27a^3}} $$

Not surprisingly, we have values for u and v.




We can then substitute the equations for $P$ and $Q$ above, and substituting for $y$ in $y-\frac{b}{3a}=x$, we get



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .