“[Zuanne de Tonini] da Coi… impuned Tartaglia to publish his method, but the latter declined to do so. In 1539 Cardan wrote to Tartaglia, and a meeting was arranged at which, Tartaglia says, having pledged Cardan to secrecy, he revealed the method in cryptic verse and later with a full explanation. Cardan admits that he received the solution from Tartaglia, but… without any explanation. At any rate, the two cubics x^3 + ax^2 = c and x^3 + bx = c could now be solved. The reduction of the general cubic x^3 + ax^2 + bx = c to the second of these forms does not seem to have been considered by Tartaglia at the time of the controversy. When Cardan published his Ars Magna however, he transformed the types x^3 = ax^2 + c and x^3 + ax^2 = c by substituting x = y + \frac{1}{3}a and x = y - \frac{1}{3}a respectively, and transformed the type x^3 + c = ax^2 by the substitution x = \sqrt[3]{c^2/y}, thus freeing the equations of the term x^2. This completed the general solution, and he applied the method to the complete cubic in his later problems.”

Source: History of Mathematics (1925) Vol.2, p.461

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David Eugene Smith33

American mathematician 1860–1944

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“The problem of the biquadratic equation was laid prominently before Italian mathematicians by Zuanne de Tonini da Coi, who in 1540 proposed the problem, "Divide 10 parts into three parts such that they shall be continued in proportion and that the product of the first two shall be 6." He gave this to Cardan with the statement that it could not be solved, but Cardan denied the assertion, although himself unable to solve it. He gave it to Ferrari, his pupil, and the latter, although then a mere youth, succeeded where the master had failed. …This method soon became known to algebraists through Cardan's Ars Magna, and in 1567 we find it used by Nicolas Petri”

David Eugene Smith (1860–1944) American mathematician

of Deventer
Source: History of Mathematics (1925) Vol.2, pp.467-468

“Vieta was the first algebraist after Ferrari to make any noteworthy advance in the solution of the biquadratic. He began with the type x^4 + 2gx^2 + bx = c, wrote it as x^4 + 2gx^2 = c - bx, added gx^2 + \frac{1}{4}y^2 + yx^2 + gy to both sides, and then made the right side a square after the manner of Ferrari. This method… requires the solution of a cubic resolvent.
Descartes (1637) next took up the question and succeeded in effecting a simple solution… a method considerably improved (1649) by his commentator Van Schooten. The method was brought to its final form by Simpson”

David Eugene Smith (1860–1944) American mathematician

1745
Source: History of Mathematics (1925) Vol.2, p.469

“Although Cardan reduced his particular equations to forms lacking a term in x^2, it was Vieta who began with the general formx^3 + px^2 + qx + r = 0and made the substitution x = y -\frac{1}{3}p, thus reducing the equation to the formy^3 + 3by = 2c.He then made the substitutionz^3 + yz = b, or y = \frac{b - z^2}{z},which led to the formz^6 + 2cz^2 = b^2,a sextic which he solved as a quadratic.”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, p.465

“Cardan's originality in the matter seems to have been shown chiefly in four respects. First, he reduced the general equation to the type x^3 + bx = c; second, in a letter written August 4, 1539, he discussed the question of the irreducible case; third, he had the idea of the number of roots to be expected in the cubic; and, fourth, he made a beginning in the theory of symmetric functions. …With respect to the irreducible case… we have the cube root of a complex number, thus reaching an expression that is irreducible even though all three values of x turn out to be real. With respect to the number of roots to be expected in the cubic… before this time only two roots were ever found, negative roots being rejected. As to the question of symmetric functions, he stated that the sum of the roots is minus the coefficient of x2</sup”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, pp.461-464

“Suppose we a certain Number of things exposed, different each from other, as a, b, c, d, e, &c.; The question is, how many ways the order of these may be varied? as, for instance, how many changes may be Rung upon a certain Number of Bells; or, how many ways (by way of Anagram) a certain Number of (different) Letters may be differently ordered?
Alt.1,21) If the thing exposed be but One, as a, it is certain, that the order can be but one. That is 1.
2) If Two be exposed, as a, b, it is also manifest, that they may be taken in a double order, as ab, ba, and no more. That is 1 x 2 = 2. Alt.3
3) If Three be exposed; as a, b, c: Then, beginning with a, the other two b, c, may (by art. 2,) be disposed according to Two different orders, as bc, cb; whence arise Two Changes (or varieties of order) beginning with a as abc, acb: And, in like manner it may be shewed, that there be as many beginning with b; because the other two, a, c, may be so varied, as bac, bca. And again as many beginning with c as cab, cba. And therefore, in all, Three times Two. That is 1 x 2, x 3 = 6.
Alt.34) If Four be exposed as a, b, c, d; Then, beginning with a, the other Three may (by art. preceeding) be disposed six several ways. And (by the same reason) as many beginning with b, and as many beginning with c, and as many beginning with d. And therefore, in all, Four times six, or 24. That is, the Number answering to the case next foregoing, so many times taken as is the Number of things here exposed. That is 1 x 2 x 3, x 4 = 6 x 4 = 24.
5) And in like manner it may be shewed, that this Number 24 Multiplied by 5, that is 120 = 24 x 5 = 1 x 2 x 3 x 4 x 5, is the number of alternations (or changes of order) of Five things exposed. (Or, the Number of Changes on Five Bells.) For each of these five being put in the first place, the other four will (by art. preceeding) admit of 24 varieties, that is, in all, five times 24. And in like manner, this Number 120 Multiplied by 6, shews the Number of Alternations of 6 things exposed; and so onward, by continual Multiplication by the conse quent Numbers 7, 8, 9, &c.;
6) That is, how many so ever of Numbers, in their natural Consecution, beginning from 1, being continually Multiplied, give us the Number of Alternations (or Change of order) of which so many things are capable as is the last of the Numbers so Multiplied. As for instance, the Number of Changes in Ringing Five Bells, is 1 x 2 x 3 x 4 x 5 = 120. In Six Bells, 1 x 2 x 3 x 4 x 5 x 6 = 120 x 6 = 720. In Seven Bells, 720 x 7 = 5040. In Eight Bells, 5040 x 8 = 40320, And so onward, as far as we please.”

John Wallis (1616–1703) English mathematician

Source: A Discourse of Combinations, Alterations, and Aliquot Parts (1685), Ch.II Of Alternations, or the different Change of Order, in any Number of Things proposed.

“The discovery of Hippocrates amounted to the discovery of the fact that from the relation
(1)\frac{a}{x} = \frac{x}{y} = \frac{y}{b}it follows that(\frac{a}{x})^3 = [\frac{a}{x} \cdot \frac{x}{y} \cdot \frac{y}{b} =] \frac{a}{b}and if a = 2b, [then (\frac{a}{x})^3 = 2, and]a^3 = 2x^3.The equations (1) are equivalent [by reducing to common denominators or cross multiplication] to the three equations
(2)x^2 = ay, y^2 = bx, xy = ab[or equivalently…y = \frac{x^2}{a}, x = \frac{y^2}{b}, y = \frac{ab}{x} ]Doubling the Cube
the 2 solutions of Menaechmusand the solutions of Menaechmus described by Eutocius amount to the determination of a point as the intersection of the curves represented in a rectangular system of Cartesian coordinates by any two of the equations (2).
Let AO, BO be straight lines placed so as to form a right angle at O, and of length a, b respectively. Produce BO to x and AO to y.
The first solution now consists in drawing a parabola, with vertex O and axis Ox, such that its parameter is equal to BO or b, and a hyperbola with Ox, Oy as asymptotes such that the rectangle under the distances of any point on the curve from Ox, Oy respectively is equal to the rectangle under AO, BO i. e. to ab. If P be the point of intersection of the parabola and hyperbola, and PN, PM be drawn perpendicular to Ox, Oy, i. e. if PN, PM be denoted by y, x, the coordinates of the point P, we shall have

\begin{cases}y^2 = b. ON = b. PM = bx\\ and\\ xy = PM. PN = ab\end{cases}whence\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.
In the second solution of Menaechmus we are to draw the parabola described in the first solution and also the parabola whose vertex is O, axis Oy and parameter equal to a.”

Thomas Little Heath (1861–1940) British civil servant and academic

The point P where the two parabolas intersect is given by<center><math>\begin{cases}y^2 = bx\\x^2 = ay\end{cases}</math></center>whence, as before,<center><math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math></center>
Apollonius of Perga (1896)