### Part I. Section IV. Chapter 12. “Of the Rule of Cardano, or of Scipione del Ferro.”

734 When we have removed fractions from an equation of the third degree, according to the manner which has been explained, and none of the divisors of the last term are found to be a root of the equation, it is a certain proof, not only that the equation lias no root in integer numbers, but also that a fractional root cannot exist; which may be proved as follows.

Let there be given the equation $$x^3-ax^2+bx-c=0$$, in which, $$a$$, $$b$$, $$c$$ express integer numbers. If we suppose, for example, $$x=\frac{3}{2}$$, we shall have $$\frac{27}{8}-\frac{9}{4}a+\frac{3}{2}b-c=0$$. Now, the first term only has 8 for the denominator; the others being either integer numbers, or numbers divided only by 4 or by 2, and therefore cannot make 0 with the first term. The same thing happens with every other fraction.

735 As in those fractions the roots of the equation are neither integer numbers, nor fractions, they are irrational, and, as it often happens, imaginary. The manner, therefore, of expressing them, and of determining the radical signs which affect them, forms a very important point, and deserves to be carefully explained. This method, called Cardano’s Rule, is ascribed to Cardano, or more properly to Scipione del Ferro, both of whom lived some centuries since.

736 In order to understand this rule, we must first attentively consider the nature of a cube, whose root is a binomial.

Let $$a+b$$ be that root; then the cube of it will be $$a^3+3a^2b+3ab^2+b^3$$, and we see that it is composed of the cubes of the two terms of the binomial, and beside that, of the two middle terms, $$3a^2b+3ab^2$$, which have the common factor $$3ab$$, multiplying the other factor, $$a+b$$; that is to say, the two terms contain thrice the product of the two terms of the binomial, multiplied by the sum of those terms.

737 Let us now suppose $$x=a+b$$; taking the cube of each side, we have $$x^3=a^3+b^3+3ab(a+b)$$: and, since $$a+b=x$$, we shall have the equation, $$x^3=a^3+b^3+3abx$$ or $$x^3=3abx+a^3+b^3$$, one of the roots of which we know to be $$x=a+b$$. Whenever, therefore, such an equation occurs, we may assign one of its roots.

For example, let $$a=2$$ and $$b=3$$; we shall then have the equation $$x^3=18x+35$$, which we know with certainty to have $$x=5$$ for one of its roots.

738 Farther, let us now suppose $$a^3=p$$ and $$b^3=q$$; we shall then have $$a=\sqrt{\vphantom{p}}p$$ and $$b=\sqrt{\vphantom{q}}q$$, consequently, $$ab=\sqrt{\vphantom{pq}}pq$$; therefore, whenever we meet with an equation, of the form $$x^3=3x\sqrt{\vphantom{pq}}pq+p+q$$, we know that one of the roots is $$\sqrt{\vphantom{p}}p+\sqrt{\vphantom{q}}q$$.

Now, we can determine $$p$$ and $$q$$, in such a manner, that both $$3\sqrt{\vphantom{pq}}pq$$ and $$p+q$$ may be quantities equal to determinate numbers; so that we can always resolve an equation of the third degree, of the kind which we speak of.

739 Let, in general, the equation $$x^3=fx+g$$ be proposed. Here, it will be necessary to compare $$f$$ with $$3\sqrt{\vphantom{pq}}pq$$ and $$g$$ with $$p+q$$; that is, we must determine $$p$$ and $$q$$ in such a manner, that $$3\sqrt{\vphantom{pq}}pq$$ may become equal to $$f$$, and $$p+q=g$$; for we then know that one of the roots of our equation will be $$x=\sqrt{\vphantom{p}}p+\sqrt{\vphantom{q}}q$$.

740 We have therefore to resolve these two equations,

$\begin{gather} 3\sqrt{\vphantom{pq}}pq=f,\\ p+q=g. \end{gather}$

The first gives $$\sqrt{\vphantom{pq}}pq=\dfrac{f}{3}$$; or $$pq=\dfrac{f^3}{27}=\frac{1}{27}f^3$$, and $$4pq=\frac{4}{27}f^3$$. The second equation, being squared, gives $$p^2+2pq+q^2=g^2$$; if we subtract from it $$4pq=\frac{4}{27}f^3$$, we have $$q^2-2pq+q^2=g^2-\frac{4}{27}f^3$$, and taking the square root of both sides, we have

$p-q=\surd\left(g^2-\frac{4}{27}f^3\right).$

Now, since $$p+q=g$$, we have, by adding $$p+q$$ to one side of the equation, and its equal, $$g$$, to the other, $$2p=g+\surd\left(g^2-\frac{4}{27}f^3\right)$$; and by subtracting $$p-q$$ from $$p+q$$, we have $$2q=g-\surd\left(g^2-\frac{4}{27}f^3\right)$$; consequently,

$p=\dfrac{g+\surd\left(g^2-\frac{4}{27}f^3\right)}{2},\quad \textrm{and} \quad q=\dfrac{g-\surd\left(g^2-\frac{4}{27}f^3\right)}{2}.$

741 In a cubic equation, therefore, of the form $$x^3=fx+g$$, whatever be the numbers $$f$$ and $$g$$, we have always for one of the roots

$x=\sqrt{\vphantom{\dfrac{g+\surd\left(g^2-\frac{4}{27}f^3\right)}{2}}}\dfrac{g+\surd\left(g^2-\frac{4}{27}f^3\right)}{2} + \sqrt{\vphantom{\dfrac{g-\surd\left(g^2-\frac{4}{27}f^3\right)}{2}}} \dfrac{g-\surd\left(g^2-\frac{4}{27}f^3\right)}{2};$

that is, an irrational quantity, containing not only the sign of the square root, but also the sign of the cube root; and this is the formula which is called the Rule of Cardano.

742 Let us apply it to some examples, in order that its use may be better understood.

Let $$x^3=6x+9$$. First, we shall have $$f=6$$, and $$g=9$$; so that $$g^2=81$$, $$f^2=216$$, $$\frac{4}{27}f^3=32$$; then $$g^2-\frac{4}{27}f^3=49$$, and $$\surd\left(g-\frac{4}{27}f^3\right)=7$$. Therefore, one of the roots of the given equation is

$x=\sqrt{\vphantom{\frac{9+7}{2}}} \frac{9+7}{2} + \sqrt{\vphantom{\frac{9-7}{2}}} \frac{9-7}{2} = \sqrt{\vphantom{\frac{16}{2}}} \frac{16}{2} + \sqrt{\vphantom{\frac{2}{2}}} \frac{2}{2} = \sqrt{\vphantom{8}} 8 + \sqrt{\vphantom{1}} 1 = 2+1=3.$

743 Let there be proposed the equation $$x^2=3x+2$$. Here, we shall have $$f=3$$ and $$g=2$$; and consequently, $$g^2=4$$, $$f^3=27$$, and $$\frac{4}{27}f^3=4$$; which gives $$\surd \left(g^2-\frac{4}{27}f^3\right)=0$$; whence it follows, that one of the roots is

$x=\sqrt{\vphantom{\frac{2+0}{2}}} \frac{2+0}{2}+\sqrt{\vphantom{\frac{2-0}{2}}} \frac{2-0}{2}=1+1=2.$

744 It often happens, however, that, though such an equation has a rational root, that root cannot be found by the rule which we are now considering.

Let there be given the equation $$x^3=6x+40$$, in which $$x=4$$ is one of the roots. We have here $$f=6$$ and $$g=40$$; farther, $$g^2=1600$$, and $$\frac{4}{27}f^3=32$$; so that $$g^2-\frac{4}{27}f^3=1568$$, and

$\surd\left(g-\frac{4}{27}f^3\right)=\surd 1568=\surd (4\cdot 4\cdot 49\cdot 2)=28\surd 2;$

consequently one of the roots will be

$x=\sqrt{\vphantom{\frac{40+28\surd 2}{2}}} \frac{40+28\surd 2}{2} + \sqrt{\vphantom{\frac{40-28\surd 2}{2}}} \frac{40-28\surd 2}{2}$

or

$x=\sqrt{\vphantom{1}} (20+14\surd 2) + \sqrt{\vphantom{1}} (20-14\surd 2);$

which quantity is really =4, although, upon inspection, we should not suppose it. In fact, the cube root of 2+√2 being 20+14√2, we have, reciprocally, the cube root of 20+14√2 equal to 2+√2; in the same manner, $$\sqrt{1}(20-14\surd 2)=2-\surd 2$$; wherefore our root $$x=2+\surd 2+2-\surd 2=4$$.

745 To this rule it might be objected, that it does not extend to all eqviations of the third degree, because the square of $$x$$ does not occur in it; that is to say, the second term of the equation is wanting. But we may remark, that every complete equation may be transformed into another, in which the second term is wanting, which will therefore enable us to apply the rule.

To prove this, let us take the complete equation $$x^3-6x^2+11x-6=0$$: where, if we take the third of the coefficient 6 of the second term, and make $$x-2=y$$, we shall have

$x=y+2,\qquad x^2=y^2+4y+4,\quad \textrm{and} \quad x^3=y^3+6y^2+12y+8;$

Consequently,

 $$x^3$$ $$y^3$$ $$+6y^2$$ $$+12y$$ +8 $$-6x^2$$ $$-6y^2$$ $$-24y$$ -24 $$11x$$ $$11y$$ +22 -6 -6 or, $$x^3-6x^2+11x-6$$ $$y^3$$ $$-y$$

We have, therefore, the equation $$y^3-y=0$$, the resolution of which it is evident, since we immediately perceive that it is the product of the factors

$y(y^2-1)=y(y+1)(y-1)=0.$

If we now make each of these factors =0, we have

I II III
$$y=0$$, $$y=-1$$, $$y=1$$,
$$x=2$$, $$x=1$$, $$x=3$$,

that is to say, the three roots which we have already found.

746 Let there now be given the general equation of the third degree, $$x^3+ax^2+bx+c=0$$, of which it is required to destroy the second term.

For this purpose, we must add to $$x$$ the third of the coefficient of the second term, preserving the same sign, and then write for this sum a new letter, as for example $$y$$, so that we shall have $$x+\frac{1}{3}a=y$$, and $$x=y-\frac{1}{3}a$$; whence results the following calculation:

$x=y-\frac{1}{3}a, \qquad x^2=y^2-\frac{2}{3}ay+\frac{1}{9}a^2,$

and

$x^3=y^3-ay^2+\frac{1}{3}a^2y+\frac{1}{27}a^3$

Consequently,

 $$x^3$$ $$y^3$$ $$-ay^2$$ $$+\frac{1}{3}a^2y$$ $$-\frac{1}{27}a^3$$ $$ax^2$$ $$ay^2$$ $$-2/3a^2y$$ $$+\frac{1}{9}a^3$$ $$bx$$ $$by$$ $$-\frac{1}{3}ab$$ $$c$$ $$c$$

or,

$y^2 - \left(\frac{1}{3}a-b\right)y + \frac{2}{27}a \sqrt{\vphantom{-\frac{1}{3}ab}} -\frac{1}{3}ab + c =0,$

an equation in which the second term is wanting.

747 We are enabled, by means of this transformation, to find the roots of all equations of the third degree, as the following example will show.

Let it be proposed to resolve the equation

$x^3-6x^2+13x-12=0.$

Here it is first necessary to destroy the second term; for which purpose, let us make $$x-2=y$$, and then we shall have $$x=y+2$$, $$x^2=y^2+4y+4$$, and $$x^3=y^3+6y^2+12y+8$$; therefore

 $$x^3$$ = $$y^3$$ $$+6y^2$$ $$+12y$$ +8 $$-6x^2$$ = $$-6y^2$$ $$-24y$$ -24 $$13x$$ = $$13y$$ +26 -12 = -12

which gives $$y^3+y-2=0$$; or $$y^3=-y+2$$.

And if we compare this equation with the formula, (Art. 741) $$x^3=fx+g$$, we have $$f=-1$$, and $$g=2$$; wherefore, $$g^2=4$$, and $$\frac{4}{27}f^3=-\frac{4}{27}$$; also, $$g^2-\frac{4}{27}f^3=4+\frac{4}{27}=\frac{112}{27}$$, and

$\surd \left(g^2-\frac{4}{27}f^3\right)= \surd \frac{112}{27} = \dfrac{4\surd 21}{9};$

consequently,

$y=\sqrt{\vphantom{\dfrac{2+\frac{4\surd 21}{9}}{2}}} \left( \dfrac{2+\frac{4\surd 21}{9}}{2} \right) +\sqrt{\vphantom{\dfrac{2-\frac{4\surd 21}{9}}{2}}} \left( \dfrac{2-\frac{4\surd 21}{9}}{2} \right),$

or

$y= \sqrt{\vphantom{1+\dfrac{2\surd 21}{9}}} \left(1+\dfrac{2\surd 21}{9} \right) + \sqrt{\vphantom{1-\dfrac{2\surd 21}{9}}} \left(1-\dfrac{2\surd 21}{9} \right),$

or

$y=\sqrt{\vphantom{\dfrac{9+2\surd 21}{9}}} \left( \dfrac{9+2\surd 21}{9} \right) + \sqrt{\vphantom{\dfrac{9-2\surd 21}{9}}} \left( \dfrac{9-2\surd 21}{9} \right),$

or

$y=\sqrt{\vphantom{\dfrac{27+7\surd 21}{27}}} \left( \dfrac{27+7\surd 21}{27} \right) + \sqrt{\vphantom{\dfrac{27-7\surd 21}{27}}} \left( \dfrac{27-7\surd 21}{27} \right),$

or

$y=\frac{1}{3}\sqrt{\vphantom{1}}(27+6\surd 21) + \frac{1}{3}\sqrt{\vphantom{1}}(27-6\surd 21);$

and it remains to substitute this value in $$x=y+2$$.

748 In the solution of this example, we have been brought to a quantity doubly irrational; but we must not immediately conclude that the root is irrational: because the binomials $$27 \pm 6\surd 21$$ might happen to be real cubes; and this is the case here; for the cube of $$\dfrac{3+\surd 21}{2}$$ being $$\dfrac{216+48\surd 21}{8}=27+6\surd 21$$, it follows that the cube root of $$27-6\surd 21$$ is $$\dfrac{3-\surd 21}{2}$$. Hence the value which we found for $$y$$ becomes

$y=\frac{1}{3}\left( \dfrac{3+\surd 21}{2}\right)+\frac{1}{3}\left( \dfrac{3-\surd 21}{2}\right)=\frac{1}{2}+\frac{1}{2}=1.$

Now, since $$y=1$$, we have $$x=3$$ for one of the roots of the equation proposed, and the other two will be found by dividing the equation by $$x-3$$. Also making the quotient $$x^2-3x+4=0$$, we have $$x^2=3x-4$$; and

$x=\frac{3}{2} \pm \surd \left(\frac{9}{4}-\frac{16}{4}\right)=\frac{3}{2} \pm \surd -\frac{7}{4} = \dfrac{3 \pm \surd -7}{2};$

which are the other two roots, but they are imaginary.

749 It was, however, by chance, as we have remarked, that we were able, in the preceding example, to extract the cube root of the binomials that we obtained, which is the case only when the equation has a rational root; consequently, the rules of the preceding chapter are more easily employed for finding that root. But when there is no rational root, it is, on the other hand, impossible to express the root which we obtain in any other way, than according to the rule of Cardano; so that it is then impossible to apply reductions. For example, in the equation $$x^3=6x+4$$, we have $$f=6$$ and $$g=4$$; so that

$x=\sqrt{\vphantom{1}}(2+2\surd -1)+\sqrt{\vphantom{1}}(2-2\surd -1),$

which cannot be otherwise expressed.

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.