# How Tartaglia Solved the Cubic Equation

Author(s):
Friedrich Katscher

Today we have one cubic equation, which we represent

 x3+nx2+px+q=0  (n,  p,  q   positive,  negative,  or  zero),

and which comprises all possible cases.  But imagine living at a time when 0 was only a digit, and not regarded as a number, and, therefore, setting equal to zero was unknown.

And imagine moreover that negative numbers, and also negative solutions of equations, were rejected - called false or fictitious - because at that time one thought geometrically, and the side of a square, or the edge of a cube cannot be negative. It was the French-Dutch mathematician Albert Girard (1595-1632), who in his 1629 book Invention nouvelle en algebre (New invention in algebra) was the first to explain the minus sign geometrically: "The solution with minus is explained in geometry by retrograding, & the minus goes back, where the plus advances." He also urged, "the solutions with minus should not be omitted." But the consequence of geometric thinking in the 16th century was that, with rare exceptions, only positive terms, only plus signs, were allowed on both sides of an equation.

That meant that, instead of the single cubic equation of today, there were at least 13 equations, 7 with all four terms (cubic, quadratic, linear, and absolute term), 3 without the linear term, and 3 without the quadratic term:

7 complete cubic equations (all powers represented):
x3+nx2+px=q
x3+nx2+q=px
x3+px+q=nx2
x3+nx2=px+q
x3+px=nx2+q
x3+q=nx2+px
x3=nx2+px+q

3 equations without the linear term:
x3+nx2=q
x3=nx2+q
x3+q=nx2

3 equations without the quadratic term:
A) x3+px=q
B) x3=px+q
C) x3+q=px

Each type of cubic equation was treated separately. But, in the early 1500s, the ten cubic equations containing the quadratic term were too difficult to be solved. At first, only the three types we call here A), B) and C) were accessible.

# How Tartaglia Solved the Cubic Equation - Cubic Equations

Author(s):
Friedrich Katscher

Today we have one cubic equation, which we represent

 x3+nx2+px+q=0  (n,  p,  q   positive,  negative,  or  zero),

and which comprises all possible cases.  But imagine living at a time when 0 was only a digit, and not regarded as a number, and, therefore, setting equal to zero was unknown.

And imagine moreover that negative numbers, and also negative solutions of equations, were rejected - called false or fictitious - because at that time one thought geometrically, and the side of a square, or the edge of a cube cannot be negative. It was the French-Dutch mathematician Albert Girard (1595-1632), who in his 1629 book Invention nouvelle en algebre (New invention in algebra) was the first to explain the minus sign geometrically: "The solution with minus is explained in geometry by retrograding, & the minus goes back, where the plus advances." He also urged, "the solutions with minus should not be omitted." But the consequence of geometric thinking in the 16th century was that, with rare exceptions, only positive terms, only plus signs, were allowed on both sides of an equation.

That meant that, instead of the single cubic equation of today, there were at least 13 equations, 7 with all four terms (cubic, quadratic, linear, and absolute term), 3 without the linear term, and 3 without the quadratic term:

7 complete cubic equations (all powers represented):
x3+nx2+px=q
x3+nx2+q=px
x3+px+q=nx2
x3+nx2=px+q
x3+px=nx2+q
x3+q=nx2+px
x3=nx2+px+q

3 equations without the linear term:
x3+nx2=q
x3=nx2+q
x3+q=nx2

3 equations without the quadratic term:
A) x3+px=q
B) x3=px+q
C) x3+q=px

Each type of cubic equation was treated separately. But, in the early 1500s, the ten cubic equations containing the quadratic term were too difficult to be solved. At first, only the three types we call here A), B) and C) were accessible.

# How Tartaglia Solved the Cubic Equation - The First Solutions

Author(s):
Friedrich Katscher

The Bolognese university lecturer Scipione dal Ferro (1465-1526) found solutions to equations of types

A) x3+px=q, B) x3=px+q, and C) x3+q=px

in 1505 (by one report) or 1515 (by another report), but did not publish them. One of his pupils, the Venetian arithmetician Antoniomaria Fior, challenged the reckoning master Nicolo Tartaglia (1499/1500-1557) to a written mathematical contest. Each one made up 30 mathematical problems, with Fior's set consisting only of equations of the type we called A), and Tartaglia's 30 different algebraic and geometric questions.

Nicolo Tartaglia (1500-1557)

Luckily Tartaglia found the "general rule," as he called it, for type A) equations on February 12, 1535, eight days before the deadline to bring the solutions to a notary, and for type B) equations on the following day. He solved Fior's problems within two hours, whereas Fior was unable to answer Tartaglia's mathematical questions.

In the 1530s algebraic quantities were not yet represented by letters. The first who did this was the Frenchman François Viète (Latinized Vieta, 1540-1603) in 1591; he used the vowels A, E, I, O, U, Y for the unknowns, and the first Greek consonants - B(eta), G(amma), D(elta), Z(eta) - for constants. The letter x for the unknown was introduced by the Frenchman René Descartes (Latinized Cartesius, 1596-1650) in 1637. Therefore, in the 1530s, you had to either write out in full the terms used, or use abbreviations to save time and space.

Tartaglia denoted the three equations without the quadratic term thus:

A) Capitolo de cubo e cosa equal à numero,

B) Capitolo de cose e numero equal à cubo,

C) Capitolo de cubo e numero equal à cose.

The Italian word capitolo has the meaning of chapter or funny poem, but the mathematicians used it for "type of equation." Cosa, plural cose, means thing and was the term for the unknown, our x, but also for the whole linear term, our px, and even for our p alone. Cubo, of course, was our x3, and numero, number, was the designation of the absolute or constant term, our q. The word de means "of", e "and", and equal à "equal to". Now it is easy to understand the wording of Tartaglia although it is puzzling why in A) he used the singular cosa and in B) and C) the plural cose.

# How Tartaglia Solved the Cubic Equation - Tartaglia's Solution

Author(s):
Friedrich Katscher

It seems possible to guess how Tartaglia discovered the solutions of the three cubic equations we called A), B) and C). He gave some hints, which suggest that the way described here is the right one.

First we have to report that Tartaglia thought he discovered what we express in the formula (a+b)3=a3+3a2b+3ab2+b3 although it had been found already by an Arabian mathematician in the 12th century.

Tartaglia published the first Italian translation of Euclid's Elements in 1543. But he had studied the work of the Greek mathematician before this time. He gave lectures about him in 1536 in Venice and knew his work well.

Euclid's Tenth Book treats irrational numbers and combinations of square roots, and distinguished between those forming a sum and those forming a difference (written in modern notation): $a+\sqrt{b}, \sqrt{a} + b, \sqrt{a} + \sqrt{b},$ called binomium in Latin (from the prefix bi-, two, and nomen, name) and binomio in Italian, and $a-\sqrt{b}, \sqrt{a} - b, \sqrt{a} - \sqrt{b},$ called residuum (remainder) or recisum (cut off) in Latin and residuo or reciso in Italian. Today both kinds are called binomials.

The quadratic equation has a solution in the form of a binomio $$a+\sqrt{b}$$ or of a residuo $$a-\sqrt{b}.$$ This suggested that cubic equations might also have a solution in the form of a binomio or of a residuo, however, with cube roots instead of quadratic ones. Probably Tartaglia tried the different possibilities, and thus found that the residuo $$\sqrt[3]{a} - \sqrt[3]{b}$$ and the binomio $$\sqrt[3]{a} + \sqrt[3]{b}$$ led to the solutions he sought.

# How Tartaglia Solved the Cubic Equation - Tartaglia's Solution in Modern Notation

Author(s):
Friedrich Katscher

We describe first, in modern notation, the solution method for equation A) x3+px=q: If you cube the residuo, the difference of two cube roots,  $$\sqrt[3]{u} - \sqrt[3]{v},$$ where at that time always $$u > v,$$ you get ${\big( {\sqrt[3]{u} - \sqrt[3]{v}}\big)}^3 = u - 3{\sqrt[3]{u}}{\sqrt[3]{v}}\big( {\sqrt[3]{u}}-{\sqrt[3]{v}}\big) - v,$ or ${\big( {\sqrt[3]{u} - \sqrt[3]{v}}\big)}^3 + 3{\sqrt[3]{uv}}\big( {\sqrt[3]{u}}-{\sqrt[3]{v}}\big) = u - v.$

If you compare this with

 x3+px=q,

$$\sqrt[3]{u} - \sqrt[3]{v}$$ corresponds to $$x,$$ $$3{\sqrt[3]{uv}}$$ to $$p,$$ and $$u-v$$ to $$q.$$ From $$3{\sqrt[3]{uv}} = p,$$ you obtain $$uv = \Big({\frac{p}{3}}\Big)^3,$$ which, when combined with $$u-v = q,$$ yields quadratic equations $$u^2 - qu - \Big({\frac{p}{3}}\Big)^3 = 0$$ and $$v^2 + qv - \Big({\frac{p}{3}}\Big)^3 = 0$$ with positive solutions: $u = \frac{q}{2} + \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}$ $v = -\frac{q}{2} + \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}$

Therefore, remembering that $$x = \sqrt[3]{u} - \sqrt[3]{v},$$ the explicit solution of equation A) in modern notation is: $x = \sqrt[3]{ \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}+\frac{q}{2}} - \sqrt[3]{ \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}-\frac{q}{2}}.$

In the case B) x3=px+q, the path to the solution begins with cubing the binomio, the sum of two cube roots: ${\big( {\sqrt[3]{u} + \sqrt[3]{v}}\big)}^3 = 3{\sqrt[3]{uv}}\big( {\sqrt[3]{u}}+{\sqrt[3]{v}}\big) + u + v.$

Comparing this with

 x3=px+q,

you have again $$uv = \Big({\frac{p}{3}}\Big)^3;$$ this time, however, $$u+v = q$$ and $$x = \sqrt[3]{u} + \sqrt[3]{v}.$$

Solving the quadratic equations you get this time: $u = \frac{q}{2} + \sqrt{\Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}$ $v = \frac{q}{2} - \sqrt{\Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}$ provided $$\Big({\frac{q}{2}}\Big)^2 \ge \Big({\frac{p}{3}}\Big)^3 .$$

Therefore, the explicit solution of equation B) in modern notation is $x = \sqrt[3]{\frac{q}{2} + \sqrt{ \Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}} + \sqrt[3]{\frac{q}{2} - \sqrt{\Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}}.$

With the same p and q, equation C) x3 + q = px has the same solution as B), however, with opposite signs. $x = -\sqrt[3]{\frac{q}{2} + \sqrt{ \Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}} - \sqrt[3]{\frac{q}{2} - \sqrt{\Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}}.$ This is so because equation C) written with opposite signs, -x3+q=-px, is identical with B), x3=px+q.

Because the negative solution of C) was rejected,this type of cubic equation was not treated by most contemporary mathematicians.

# How Tartaglia Solved the Cubic Equation - Tartaglia's Description of His Solution

Author(s):
Friedrich Katscher

We will show here how Tartaglia solved the equation

.1.cubo piu .3.cose, equal à .10.

(x3+3x = 10; p=3, q=10; piu, today written with an accent, più, means plus; numbers were always between dots). He wrote on April 23, 1539 in a letter to M. (Messer, Mr.) Hieronimo Cardano that

you will have to find two numbers that the difference between the two is .10. (that is, so much as is our number) & that we make the product of these two quantities, the one multiplied by the other, exactly .1., that is, the cube of the third part of the cose,...

This means in our modern notation that first we have to find the two numbers u and v that fulfill u-v=q=10 and uv=(p/3)3=(3/3)3=1.

Now Tartaglia assumed that the mathematician knew how to solve such a quadratic equation. He continued:

which two numbers or quantities operating by Algebra, or by some other way, which seems more comfortable, you will find the one of them, namely the smaller one, R.26.men.5., & the other, that is, the larger one, R.26.piu.5.

Here, R (radix, Latin, radice, Italian) is the symbol for square root, R.cuba, abbreviated R.cu., for cube root. This means Tartaglia found $v = \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}-\frac{q}{2}$ $= \sqrt{\Big({\frac{10}{2}}\Big)^2 + \Big({\frac{3}{3}}\Big)^3}-\frac{10}{2}$ $= \sqrt{25+1} - 5 = \sqrt{26}-5 ,$ and in the same way $$u=\sqrt{26}+5.$$

Now it is necessary to find from each of these two quantities its lato cubico, that is, its R.cuba, & that of the smaller one will be R.universale cuba de R.26.men.5 & that of the larger one will be [R.universale cu.R.26.pui.5], and its remainder will be the value of our cosa principale, which remainder will be the difference of these two R.universale cu., that is it will be

R.u.cu.R.26.piu.5.men.R.u.cuba R.26.men.5.

& so much is the value of our cosa principale...

Here, the "value of our cosa principale" is $$\sqrt[3]{\sqrt{26}+5} - \sqrt[3]{\sqrt{26} - 5}.$$ Also, R.universale, abbreviated R.u., means the root of the whole following expression, so that, for instance, "that of the smaller one" is $$\sqrt[3]{\sqrt{26} - 5}$$.

It is typical that the results in most cases were left like this; that is, the square and cube roots were not extracted. Today we have the final result as the decimal number x1=1.698885... (x2 and x3 are complex conjugates). The decimal numbers were introduced only in 1585 by the Dutchman Simon Stevin (1548-1620), but, of course, before that there were approximations of square and cube roots with ordinary fractions.

# How Tartaglia Solved the Cubic Equation - Tartaglia's Poem

Author(s):
Friedrich Katscher

Tartaglia divulged to Hieronimo Cardano (1501-1576) the solution of the three cubic equations without the quadratic term on March 25, 1539 in Cardano's house in Milano in the form of a famous poem (translated here by the author):

01) When the cube with the cose beside it <x3+px >
02) Equates itself to some other whole number, <=q>
03) Find two other, of which it is the difference. <u-v=q>
04) Hereafter you will consider this customarily
05) That their product always will be equal <uv=>
06) To the third of the cube of the cose net. 3/3, instead of (p/3)3>
07) Its general remainder then
08) Of their cube sides , well subtracted, <$${\sqrt[3]u}-{\sqrt[3]v}$$>
09) Will be the value of your principal unknown. <=x>
10) In the second of these acts,
11) When the cube remains solo , <x3=px+q>
12) You will observe these other arrangements:
13) Of the number <q> you will quickly make two such parts, <q=u+v>
14) That the one times the other will produce straightforward <uv=>
15) The third of the cube of the cose in a multitude, <p3/3, instead of (p/3)3>
16) Of which then, per common precept,
17) You will take the cube sides joined together. ﻿﻿<$${\sqrt[3]u}+{\sqrt[3]v}$$﻿﻿ ﻿﻿﻿
18) And this sum will be your concept. <=x>
19) The third then of these our calculations <x3+q=px>
20) Solves itself with the second, if you look well after,
21) That by nature they are quasi conjoined.
22) I found these, & not with slow steps,
23) In thousand five hundred, four and thirty
24) With very firm and strong foundations
25) In the city girded around by the sea.

Of course, in the English translation the lines do not rhyme as they do in the Italian original. There the poem is made highly artistic. Tartaglia chose a kind of poem - a so-called capitolo (the same word also means chapter, and type of equation) - constructed in a very complicated manner. A capitolo poem is written in the so-called terza rima (Italian: third rhyme), the structure of which consists of groups of three lines of verse (called tercet in English), lines 01 to 03, 04 to 06, 07 to 09, etc., where the first and the third line rhyme. The peculiarity of this kind of poems is that the middle line of a tercet rhymes with the first and the third line of the next triplet. If the lines are represented by letters, and the same letter means that these lines rhyme, the pattern of a terza rima is aba, bcb, cdc, ded, and so on. (The poem in the original Italian appears on page 8.)

The necessity to follow this scheme led to two misleading instructions in line 6 and 15. In line 6 the wording says "the third of the cube of the cose net." The "cose net" in this case is the coefficient p. "The third of the cube," therefore, means p3/3. That is wrong. Correctly it should be "the cube of the third of the unknown," (p/3)3, and that is why this passage confused Cardano. In a letter from 9 April he implored Tartaglia for help to solve the equation .1.cubo piu.3.cose equal à 10. Tartaglia answered on April 23, 1539, and gave the explanatory instructions we describe on page 5.

Lines 19, 20 and 21 mean that you solve this type of cubic equation, our C), with the second type B) x3=px+q. The second type x3=px+q and the third type x3+q=px with the same p and q, as was said before, have numerically the same three roots but with opposite signs. Type B) always has a positive root x1. Consequently, the root of the corresponding equation of the third type was negative, and, therefore, rejected.

Line 23: In the Republic of Venice until its end in 1797 the year did not begin on January 1 as today but officially on March 1. Therefore, February 12 and 13 of our year 1535 were still in the Venetian year 1534.

# How Tartaglia Solved the Cubic Equation - The Other Cubic Equations

Author(s):
Friedrich Katscher

To solve the ten cubic equations with the quadratic term, this term has to be removed by transforming it into a linear term, and the transformed equation then can be solved by the formula of Scipione dal Ferro and Tartaglia, which today is unjustly called Cardano's formula because he was the first to publish it, in his Ars magna in 1545.

Tartaglia did not find a way to solve the ten cubic equations with the quadratic term. It is true that he printed the solution of the equation .1.cubo piu .6.censi equal à .100. (x3+6x2=100) in his book Quesiti et inventioni diverse (Diverse problems and inventions) of 1546, but this example was plagiarized from Chapter XV of Cardano's Ars magna of 1545, and he even introduced an error.

Cardano gave the solution in Latin:

RV:cubica 42 p:R 1700 p:RV:cubica 42 m:R  1700 m:2,

where V is the abbreviation of Vniversalis, p: means plus, and m: minus. Therefore, the root in modern notation is ${\sqrt[3]{42+\sqrt{1700}}}+{\sqrt[3]{42-\sqrt{1700}}}-2\,\,(=3.282260\dots).$

But Tartaglia's solution

R.u.cu.42.piu R.17000. piu R.u.cu.42.men.R.17000.men.2.

twice has 17000 instead of the correct 1700.

Scipione dal Ferro and Nicolo Tartaglia found the solution of the three cubic equations without a quadratic term.

But it was Cardano's great immortal feat to have solved the other ten cubic equations with a quadratic term, and thus to lead the way to a general solution.

# How Tartaglia Solved the Cubic Equation - Tartaglia's Original Poem

Author(s):
Friedrich Katscher

Dapoi terrai questo per consueto
Che'llor produtto sempre sia eguale
Alterzo cubo delle cose neto,

In el secondo de cotestiatti
Quando
che'l cubo restasse lui solo
Tu
osseruarai quest'altri contratti,

Del numer farai due tal part'à uolo
Che
l'una in l'altra si produca schietto
El terzo cubo delle cose in stolo

Delle qual poi, per communprecetto
Torrai
li lati cubi insieme gionti
Et
cotal somma sara il tuo concetto.

Questi trouai, & non con paßi tardi
Nel
mille cinquecentè, quatroe trenta
Con
fondamenti ben sald'è gagliardi

Nella citta dal mar'intorno centa.

For an English translation of this poem, see page 6.