“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”

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

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David Eugene Smith33

American mathematician 1860–1944

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“[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.”

David Eugene Smith (1860–1944) American mathematician

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

“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

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David Eugene Smith33

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