In his Ars Magna Cardano specifies procedures to "depress" a cubic - a means to convert an equation such as $x^3+6x^2=100$ to $y^3=84+12y$, eliminating the $x^2$ term.

Was he the one who discovered how to do this? (Or was he just the first to publish it? (Or neither?))

In his Quesiti Tartaglia claimed to have known how to solve cubics of the form $x^3+ax^2=c$, which made me wonder if he knew how to eliminate the $x^2$ term. So far I haven't found anything definitive that indicates who first discovered the appropriate substitution.

  • 3
    $\begingroup$ The systematic use of substitutions is generally ascribed to Cardano, he showed more generally how to reduce general cubic to a depressed one. What Tartaglia did or did not know we will never know, he chose to keep it a secret for mathematical duels ("cartels") they were having at the time. All we know is that he shared how to solve the depressed cubic with Cardano, and the latter promised not to published it. The promise he did not feel like keeping after finding out that del Ferro already knew it before Tartaglia. $\endgroup$ – Conifold May 23 '19 at 2:34
  • $\begingroup$ @Conifold Weren't these sort of change of variable known to Diophantus, and possibly Archimedes or even the Babylonians? In general, they knew the effect of $x\mapsto x+a$ and $x\mapsto\lambda x$ (not in modern notation of course). See Babylonian mathematics. $\endgroup$ – Chrystomath May 24 '19 at 6:24
  • $\begingroup$ Not really. Babylonians simply gave practical instructions for manipulating numbers, and Diophantus has a single symbol for a variable, everything else is numerical. So they had no means for tracing such effects. Transformation of equations as such, even expressed verbally, is a relatively late development, see e.g. Netz From Problems to Equations. $\endgroup$ – Conifold May 24 '19 at 7:22
  • $\begingroup$ Transformation of equations is the main impetus of Al Khwarizmi's Algebra. The Babylonian tablet YBC4669 shows verbally how to convert the cubic $12x^3+x=33;22$ into $(6x)^3 + 3(6x) = 10,0,36$. $\endgroup$ – Chrystomath May 25 '19 at 6:31
  • $\begingroup$ For whatever it's worth, J. Stedall in her book From Cardano's great art to Lagrange's reflections states that Cardano was the first to transform "an equation by an operation on the roots." This is based on Cardano's example in Ars Magna where the equations $x+y=x^2$ and $xy=8$ lead to, depending on the substitution chosen, $x^2+8=x^3$ or $8y+y^3=64$. From this Cardano realized he could use a substitution to eliminate the $x^2$ term. Evidently (per Stedall) this is the first example of doing such a thing. $\endgroup$ – Brant Jul 20 '19 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.