“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

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

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David Eugene Smith 33

American mathematician 1860–1944

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“In the work of Vieta the analytic methods replaced the geometric, and his solutions of the quadratic equation were therefore a distinct advance upon those of his predecessors. For example, to solve the equation x^2 + ax + b = 0 he placed u + z for x. He then hadu^2 + (2z + a)u +(z^2 + az + b) = 0.He now let 2z + a = 0, whence z = -\frac{1}{2}a,and this gaveu^2 - \frac{1}{4}(a^2 - 4b) = 0.
u = \pm \frac{1}{2} \sqrt{a^2 - 4b}.andx = u + z = -\frac{1}{2}a \pm \sqrt{a^2 - 4b}.</center”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, p.449

“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)

“[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

“The Rule of Three, or Golden Rule of Arithmeticall whole Numbers. Be the three termes given 2 3 4. …To finde their fourth proporcionall Terme: that is to say, in such Reason to the third terme 4, as the second terme 3, is to the first terme 2 [Modern notation: \frac{x}{4} = \frac{3}{2}]. …Multiply the second terme 3, by the third terme 4, & giveth the product 12: which dividing by the first terme 2, giveth the Quotient 6: I say that 6 is the fourth proportional terme required.”

Simon Stevin (1548–1620) Flemish scientist, mathematician and military engineer

Disme: the Art of Tenths, Or, Decimall Arithmetike (1608)

“The value of the intrinsic approach is especially apparent in considering 3-dimensional congruence spaces… The intrinsic geometry of such a space of curvature K provides formulae for the surface area S and the volume V of a "small sphere" of radius r, whose leading terms are 3)S = 4 \pi r^2 (1 - \frac{Kr^2}{3} + …),
V = \frac{4}{3} \pi r^3 (1 - \frac{Kr^2}{5} + …).”

Howard P. Robertson (1903–1961) American mathematician and physicist

Geometry as a Branch of Physics (1949)

“Lety5 - ay4 + by3 - cy2 + dy - e = 0be the general equation of the fifth degree and suppose that it can be solved algebraically,—i. e., that y can be expressed as a function of the quantities a, b, c, d, and e, composed of radicals. In this case, it is clear that y can be written in the formy = p + p1R1/m + p2R2/m +…+ pm-1R(m-1)/m,m being a prime number, and R, p, p1, p2, etc. being functions of the same form as y. We can continue in this way until we reach rational functions of a, b, c, d, and e. [Note: main body of proof is excluded]
…we can find y expressed as a rational function of Z, a, b, c, d, and e. Now such a function can always be reduced to the formy = P + R1/5 + P2R2/5 + P3R3/5 + P4R4/5, where P, R, P2, P3, and P4 are functions or the form p + p1S1/2, where p, p1 and S are rational functions of a, b, c, d, and e. From this value of y we obtainR1/5 = 1/5(y1 + α4y2 + α3y3 + α2y4 + α y5) = (p + p1S1/2)1/5,whereα4 + α3 + α2 + α + 1 = 0.Now the first member has 120 different values, while the second member has only 10; hence y can not have the form that we have found: but we have proved that y must necessarily have this form, if the proposed equation can be solved: hence we conclude that
It is impossible to solve the general equation of the fifth degree in terms of radicals.
It follows immediately from this theorem, that it is also impossible to solve the general equations of degrees higher than the fifth, in terms of radicals.”

Niels Henrik Abel (1802–1829) Norwegian mathematician

A Memoir on Algebraic Equations, Proving the Impossibility of a Solution of the General Equation of the Fifth Degree (1824) Tr. W. H. Langdon, as quote in A Source Book in Mathematics (1929) ed. David Eugene Smith