We all know, a general cubic equation is of the form

$$ax^3+bx^2+cx+d=0$$ where $$a\neq0.$$

It can be easily solved with the following simple substitutions:

$$x\longmapsto x-\frac{b}{3a}$$

We get,

$$x^3+px+q=0$$ where, $p=\frac{3ac-b^2}{3a^2}$ and $q=\frac{2b^3-9abc+27a^2d}{27a^3}$

Then, using the Vieta substitution,

$$x\longmapsto x-\frac{p}{3x}$$

We get,


which is can easily turn into a quadratic equation, using the substitution: $x^3 \longmapsto x.$

And here is my question:

In mathematics is there a substitution that is "different" from the substitution $x\longmapsto x-\frac{p}{3x}$ that can be used for the standard form cubic equation $x^3+px+q=0$ , which is can easily turn into a quadratic equation?

I'm curious, if there's a new substitute I don't know about.

Thank you!

  • 2
    $\begingroup$ How different is "different"? For example, you could try $x = t/3 - p/t$. But it's only a slightly disguised version of the "Vieta" substitution. $\endgroup$ Sep 1, 2019 at 21:53
  • $\begingroup$ @Robert Israel In fact $x/3 -p/x$ is different from $x-p/(3x)$ am I wrong? $\endgroup$
    – Learner
    Sep 1, 2019 at 21:59
  • $\begingroup$ @Robert ISrael Can I ask How did you find this substitution? I can not find this in wikipedia..and Wolfram mathworld $\endgroup$
    – Learner
    Sep 1, 2019 at 22:09
  • 1
    $\begingroup$ If $x = s - p/(3s)$ then $x = at - p/(3at)$ where $s=at$. I just used $a=1/3$. $\endgroup$ Sep 2, 2019 at 1:41
  • 2
    $\begingroup$ Some older text books use the form $x^3+3Gx+2H=0$ for practical calculation as it saves a lot of fractions in the formulae. $\endgroup$ Sep 3, 2019 at 18:43

6 Answers 6


The cubic is sufficiently venerable and sufficiently simple that there are numerous other ways to solve it. Here are three of the more notable; others can be found in, e.g., Galois' Theory of Algebraic Equations by J.-P. Tignol.

Tschirnhaus Substitution

The idea is to do a more sophisticated substitution for $x$ than the linear one that removes the quadratic term: we can suppose that we start in the form $$ x^3 + px + q = 0 , $$ as it makes the calculations simpler. Since a linear substitution can be used to remove one term, one may suspect that a more general substitution can be used to remove more terms, and in particular, if we put $ y = \alpha x^2 + \beta x + \gamma $, we might hope to choose the coefficients so that both the linear term and the quadratic term vanish at the same time. This turns out to be the case, but first we must eliminate $x$ from the two equations. We have $$ \begin{align} (y-\gamma)^2 &= x^2 (\alpha x + \beta)^2 = (\alpha^2 x + 2\alpha\beta) x^3 + \beta^2 x^2 \\ &= -(\alpha^2 x + 2\alpha\beta) (px+q) + \beta^2 x^2 \\ &= (-p\alpha^2 + \beta^2)x^2 - \alpha(q\alpha + 2p\beta )x -2\alpha\beta q \end{align}$$ and $$ \begin{align} (y-\gamma)^3 &= x^3(\alpha x+\beta)^3 \\ &= -(px+q)(\alpha^3 x^3 + 3\alpha^2 \beta x^2 + 3\alpha \beta^2 x + \beta^3 ) \\ &\vdots \\ &= ( p^2\alpha^2 - 3q\alpha\beta - 3p\beta^2 ) \alpha x^2 + ( 2pq\alpha^3+ 3p^2 \alpha^2\beta - 3q\alpha\beta^2 - p\beta^3 ) x + (q\alpha^3+3p\alpha^2 \beta - \beta^3) \end{align}$$ A straightforward calculation then shows that $$ 0 = (y-\gamma)^3 + 2p\alpha (y-\gamma)^2 + (p^2\alpha^2+3q\alpha\beta + p\beta^2 ) + q(-q\alpha^3 + p\alpha^2\beta -\beta^3) , $$ and then we can expand the brackets to get $$ 0 = y^3 + (2p\alpha-3\gamma) y^2 + ( p^2\alpha^2 + 3q\alpha\beta + p\beta^2 -4p\alpha\gamma+3\gamma^2 ) y + \text{const.}, $$ where the constant is both ugly and unimportant. In particular, we can force the $y^2$ term to vanish by taking $\gamma = 2p\alpha/3$. This then leaves $$ 0 = y^3 + ( -\tfrac{1}{3} p^2\alpha^2 + 3q\alpha\beta + p\beta^2 ) y + \text{const.} $$ At this point we see that we lose nothing by taking $\alpha=1$, and thus the solution of the cubic has been reduced to the solution of the equations $$ -\tfrac{1}{3} p^2 + 3q\beta + p\beta^2 = 0 \\ y^3 + \text{const.} = 0 \\ x^2 + \beta x + \tfrac{2}{3}p - y = 0 , $$ all of which are at worst quadratic or just require finding a cube root.

Bézout's Method

This method was first considered by Bézout, and in a slightly different form by Euler. The idea is that by eliminating $z$ between the equations $$ x = a_0 + a_1 z + \dotsb + a_{n-1} z^{n-1} \\ z^n = 1 , $$ we obtain an equation of degree $n$, and the idea is to make it the right equation of degree $n$. But we know the solutions to $z^n=1$, namely the $n$th roots of unity. So eliminating $z$, the remaining equation can be written in the form $$ \prod_{\omega} (x-(a_0+a_1 \omega + \dotsb + a_{n-1}\omega^{n-1})) = 0 . $$ It remains to choose the $a_i$ appropriately. Of course the dream is to do this for any equation, but for degree higher than $4$, the degree of the equations one needs to solve becomes higher than that of the original equation, so is a complete nonstarter. But for cubics, we can manage: if $$ x^3 + bx^2+cx+d = 0 $$ is the equation, we have $$ 0 = ( x - (a_0+a_1+a_2) )( x - (a_0+a_1 \omega + a_2 \omega^2) )( x - (a_0+a_1 \omega^2+a_2 \omega) ), \tag{1} $$ where $\omega$ is now a fixed primitive cube root of unity. Expanding and using that $1+\omega+\omega^2=0$, we obtain $$ (x-a_0)^3 - 3a_1 a_2 (x-a_0) -a_1^3 -a_2^3 = 0 . $$ or $$ x^3 - 3a_0 x^2 + 3(a_0^2 - a_1 a_2) x +(- a_0^3 - a_1^3 - a_2^3 + 3a_0 a_1 a_2) = 0 $$ We now solve the equations $$ -3a_0 = b \\ 3(a_0^2 - a_1 a_2) = c \\ - a_0^3 - a_1^3 - a_2^3 + 3a_0 a_1 a_2 = d ; $$ the first is linear and gives $a_0=b/3$, and the other two become $$ a_1 a_2 = \frac{b^2-3c}{9} \\ a_1^3 + a_2^3 = d + \frac{2}{27} b^3 - \frac{bc}{3} , $$ which when one of the unknowns is eliminated becomes a bicubic equation for the other.

This is not surprising: we know that if $a_1$ is a solution, so are $\omega a_1$ and $\omega^2 a_1$, which means that $a_1^3$ has fewer possible values than $a_1$. This is key in the next method.

Lagrange Resolvents

Lagrange develops the first general ideas about how solvability by radicals should be built up, by observing how various methods, including those given above, work. Key in the theory is what happens to expressions containing the roots when the roots themselves are permuted, and it is this that eventually leads to what we now call Galois theory (in its original incarnation as the most general method for understanding the structure of solutions of equations using permutations of the roots).

For the sake of simplicity, take the same cubic equation as the previous section. Notice that if the roots are $x_1,x_2,x_3$, we can specify from $(1)$ that $$ x_1 = a_0 + a_1 + a_2 \\ x_2 = a_0 + \omega a_1 + \omega^2 a_2 \\ x_3 = a_0 + \omega^2 a_1 + \omega a_2 , $$ and so the $a_i$ can be expressed in terms of the $x_j$, as $$ a_i = \frac{1}{3} \sum_{j=1}^3 \omega^{-ij} x_j , $$ by taking linear combinations of the equations.

Suppose now that we relabel the roots. What happens to the $a_i$? There are $6$ possible ways to label the roots, so each $a_i$ can take at most $6$ values. But they certainly don't all take this many: we have $$ 3a_0 = x_1 + x_2 + x_3 = -b $$ by one of Vieta's formulae. On the other hand, a straightforward calculation shows that $a_1$ does generally have $6$ different values.

Lagrange proves that if an expression has $n$ different values under all the permutations of the roots, it is the root of an equation of degree $n$, the coefficients of which are rational functions of the coefficients of the original equation. So $a_1$ is the root of an equation of degree $6$. On the other hand, $a_1^3$ takes only $2$ different values under permutations of the roots: notice that $$ (x_1+\omega x_2+ \omega^2 x_3)^3 = (x_2+ \omega x_3 + \omega^2 x_1)^3 = ( x_3 + \omega x_1 + \omega^2 x_2 )^3 \\ (x_2+\omega x_1+ \omega^2 x_3)^3 = (x_1+ \omega x_3 + \omega^2 x_2)^3 = ( x_3 + \omega x_2 + \omega^2 x_1 )^3 , $$ using $\omega^3=1$. Therefore Lagrange tells us that $a_1^3$ is a root of a quadratic equation, namely $$ (y-(\tfrac{1}{3}(x_1+\omega x_2+ \omega^2 x_3))^3)(y-(\tfrac{1}{3}(x_2+\omega x_1+ \omega^2 x_3))^3) = 0 , $$ which we can show is $$ y^2 + \tfrac{1}{27}(2b^3-9bc+27d)y + \tfrac{1}{729}(b^2-3c)^3 = 0 \tag{2} $$ using the rest of Vieta's formulae for a cubic. This is the same quadratic that keeps appearing!

Finally, Lagrange proves an even stronger result: that if $V$ is an expression in the roots that does not change its value under a certain subset of the permutations of the roots, and $U$ is an expression that takes $m$ values under this subset, then $U$ is a root of an equation of degree $m$ with coefficients rational functions of $V$ and the coefficients of the equation. This is rather more than we need here, because if we take $V = (x_1+ \omega x_2 + \omega^2 x_3 )^3$, then we know that $ U = (x_1+ \omega x_2 + \omega^2 x_3)/3 $ has three different values under the permutations that fix $V$, namely $\sqrt[3]{V}, \omega \sqrt[3]{V} $ and $\omega^2 \sqrt[3]{V}$, obviously the solutions to $$ u^3 = V . $$

But now we know that only one permutation fixes $U$, namely the identity, and Lagrange's result implies that $x_1$ is the root of an equation of degree $1$ with coefficients rational functions of $U$ and the original equation's coefficients; i.e. a rational function of $U,b,c,d$. In particular, $$ x_1 = -\frac{b}{3} + U + \frac{b^2-3c}{9U} , $$ the other roots coming from other values of $U$.

In a certain sense this process, of finding expressions that take only a few values under certain sets of permutations, is the only way to solve equations algebraically: one can do a similar analysis of Tschirnhaus's method, although it's a lot messier. But in the case of the cubic, there is essentially only one way to do this, the way we have just given. This means that any algebraic (as opposed to transcendental, using trigonometric or hyperbolic functions) calculation of the roots will go via the quadratic $(2)$ at some point.


Not exactly what you're asking for but the substitution $x=2\sqrt{-p/3}\cos\theta$ turns $x^3+px+q=0$ into $4\cos^3\theta-3\cos\theta-3q=0$. Since $4\cos^3\theta-3\cos\theta=\cos3\theta$, this is $\cos3\theta=3q$, so $\theta=(1/3)\arccos(3q)$, and $x=2\sqrt{-p/3}\cos((1/3)\arccos(3q))$.

If $p>0$ and you don't want imaginaries then you can start with the analogous "triple angle" formula for the hyperbolic cosine, instead.


Before Viéte, there was Cardano. Suppose you have already depressed the cubic a la Tartaglia, so that it now has the form $x^3+ax+b=0,$ write this as $$x^3=px+q$$ and find $u,v$ satisfying $p=3uv$ and $q=u^3+v^3.$ This then gives you a quadratic whose roots are $u^3,v^3,$ whence you can find $u,v$ and a solution of the original cubic is $x=u+v.$ This gives the well-known formula of Cardano. But the process is easier to remember, obviously.


Here is my patent for doing the job. I found it in the school some almost 40 years ago, because i could never remember the "winning substitution" with the target of being prepared for the Olympiads. The idea is to reduce with a linear simple substitution to the equation $$ x^3+px+q=0\ .\qquad(*) $$ Now recall the formula, my favorite formula at those times: $$ x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\ . $$ Or even better, but i was not so far, since we need complex numbers and one of the primitive third roots of unity, let us write it like $\varepsilon =\frac 12(-1+\sqrt{-3})$ and delegate the choice of $\sqrt{-3}$, $$ x^3+y^3+z^3-3xyz=(x+y+z)(x+\varepsilon y+\varepsilon^2 z)(x+\varepsilon^2 y+\varepsilon z)\ . $$ And we try of course to bring $(*)$ in the above form by searching for some two values $y,z$ such that $$ \left\{ \begin{aligned} p &= -3yz\ ,\\ q &= y^3+z^3\ . \end{aligned} \right. $$ From the two equations above, we get immediately the sum and the product of the unknown quantities $y^3$ and $z^3$, this leads (Vieta) to an equation of second degree with these quantities as roots, we get $y$ by a human choice of the cubic root of $y^3$, then this determines by $p=-3yz$ the correlated cubic root of $z$ to be taken.

Now the roots are not only easily extracted, but their "common structure" is better prepared for a Galois theoretical study, they correspond to the vanishing of the one or the other factors above, so we solve for $x$ in: $$ \begin{aligned} 0 &= x+y+z\ ,&&\text{ or }\\ 0 &= x+\varepsilon y+\varepsilon^2 z\ ,&&\text{ or }\\ 0 &= x+\varepsilon^2 y+\varepsilon z\ . \end{aligned} $$

The above road map is of course not different in the result, although it is different in the exposition, but it was for me differently different, say psychologically and mnemotechnically, because it was a simple idea behind the needed substitution. And this is still the method i used didactically many times.

  • 2
    $\begingroup$ The story is longer, but for short, i wanted in school (being 14yo maybe) to understand and solve quickly and even "compose my own" problems of the shape: Show the relation: $\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=1$. The solution was to denote by $y,z$ (to get closer to the above, let us use $y,z$) the values for $\sqrt[3]{2\pm\sqrt 5}$, then to observe that $y^3+z^3$ and $yz$ can be easily computed. Playing around, a table of formulas showing me Cardano / Tartaglia, getting the equation of third degree in the foreground, and trying to get easy solutions was the door to the above story. $\endgroup$
    – dan_fulea
    Sep 1, 2019 at 23:57

There is a special case, monic and integer coefficients, when the discriminant is a perfect square. In this case, the methods of Gauss give the roots as sums of cosines. A modern treatment in COX and many, many worked examples at REUSCHLE

In the first example below, the three roots of $x^3 + x^2 - 2x-1$ are $$ 2 \cos \left( \frac{2 \pi}{7} \right) \; , \; 2 \cos \left( \frac{4 \pi}{7} \right) \; , \;2 \cos \left( \frac{8 \pi}{7} \right) \; . \; $$

In the second example below, the three roots of $x^3 + x^2 - 4x+1$ are $$ 2 \cos \left( \frac{2 \pi}{13} \right) + 2 \cos \left( \frac{10 \pi}{13} \right) \; , \; 2 \cos \left( \frac{4 \pi}{13} \right) + 2 \cos \left( \frac{6 \pi}{13} \right) \; , \; 2 \cos \left( \frac{8 \pi}{13} \right) + 2 \cos \left( \frac{12 \pi}{13} \right) \; . \; $$

  x^3 + x^2   - 2 x - 1   p  7 p.root  3 exps 6^k  d = 7^2
  x^3 + x^2   - 4 x + 1   p  13 p.root  2 exps 5^k  d = 13^2
  x^3 + x^2   - 6 x - 7   p  19 p.root  2 exps 8^k  d = 19^2
  x^3 + x^2   - 10 x - 8   p  31 p.root  3 exps 15^k  d = 2^2 * 31^2
  x^3 + x^2   - 12 x + 11   p  37 p.root  2 exps 8^k  d = 37^2
  x^3 + x^2   - 14 x + 8   p  43 p.root  3 exps 2^k  d = 2^2 * 43^2
  x^3 + x^2   - 20 x - 9   p  61 p.root  2 exps 8^k  d = 3^2 * 61^2
  x^3 + x^2   - 22 x + 5   p  67 p.root  2 exps 3^k  d = 3^2 * 67^2
  x^3 + x^2   - 24 x - 27   p  73 p.root  5 exps 7^k  d = 3^2 * 73^2
  x^3 + x^2   - 26 x + 41   p  79 p.root  3 exps 12^k  d = 79^2
  x^3 + x^2   - 32 x - 79   p  97 p.root  5 exps 19^k  d = 97^2
  x^3 + x^2   - 34 x - 61   p  103 p.root  5 exps 3^k  d = 3^2 * 103^2  
  • $\begingroup$ Any cubic equation with rational coefficients can be put into this form, so it may be slightly more general than you thought. Indeed, it's an important special case since most of the tine we deal with rationals and rational approximations to irrationals. $\endgroup$
    – Allawonder
    Sep 3, 2019 at 12:46

Possible substitution $x\to X+\frac{1}{z}$, where $X$ is some parameter and $z$ is new unknown, but additional need use some properties of cubic polynomial, which historically we received from Lagrange.

Thus we have solution of equation $f=ax+bx^2+cx^3$:

$\begin{cases} X=\dfrac{-(a b + 9 c f) + \sqrt{(a b + 9 c f)^2-4 (b^2 - 3 a c) (a^2 + 3 b f)}}{2 (b^2 - 3 a c)}\\ C=a X + b X^2 + c X^3 - f\\ B_1=a + 2 b X + 3 c X^2\\ W=B_1^3-27cC^2\\ B_2=\left\{W^{1/3}\,,-(-1)^{1/3} W^{1/3}\,,(-1)^{2/3}W^{1/3}\right\}\\ x=X+\frac{3C}{B_2-B_1} \end{cases}$

Deriving this solution see in Intro, and there is a code to verifing in Wolfram and Geogebra.


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