Tartaglia and his method

In the year 1530 Fiore challenged a mathematician from Brescia, Tonioni da Coi into a typical mathematical contest. What it was all about was to solve a certain number of problem during a limited number of days. The problems put forward by Fiore were naturally cubic equations.

Da Coi did not manage to solve these problems and turned to Tartaglia to get help. He was although not able to help him in his matter.

In 1526 Tartaglia had discovered a way of solving one kind of cubics, namely ax³ + bx² =c but this did not do as Fiores problems all were of one kind namely: ax³ +bx =c.

In February 1535 Fiore did challenge Tartaglia as he had heard rumors about him being able to solve cubic equations.

They gave each other 30 problems to solve within 40 days. The problems they gave each other were of course of the kind they managed themselves.

One of the problems Fiore gave to Tartaglia:

Which is the number that added to its cubic root becomes 6?

If the root is x the number becomes x³ and in modern notation the equation can be written as     x³ + x = 6.

Solution: we are looking for u and v  where this shall hold:     uv = 1/3    and  u³ - v³ = 6

Computing of     v                           and    u

Perform the substitution v = 1/3u in u3 - v3 = 6	Perform the substitution u = 1/3v in u3 - v3 = 6
u3 - ( 1/3u )3 = 6	( 1/3v )3 - v3 = 6
u3 - ( 1/27u 3 ) = 6	( 1/27v3 ) - v3 = 6
Multiply both sides by  27u3	Multiply both sides by  27v3
27u6 - 1 = 162u3	1 - 27v6 = 162v3
27u6 - 162u3 - 1 = 0	0 = 27v6 + 162v3 - 1
This is now a equation of second degree in u3	This is now a equation of second degree in   v3
u3 = [-(-162)+ ((-162)2-4(27)(-1))] / 54	v3 = [-(162) + ((162)2 - 4(27)(-1))] / 54
u3 = [ 162 +(26352) ] / 54	v3 = [ 162 + (26352) ] / 54
u3 = 3 + ( (26352) / 54 )	v3 = 3 + ( (26352) / 54 )
u = 3 ( [(26352) / 54 ] + 3)	v = 3 ( [(26352) / 54 ] - 3)

So   x = u - v = 3 ( [(26352) / 54 ] + 3) - 3 ( [(26352) / 54 ] - 3) = 1.634365293

Control: 1.6343652933 +1.634365293=6

When 32 days had gone Tartaglia found the method he needed and was able to solve all the 30 problems. Fiore, who not had succeeded to solve one single problem was declared as the looser.

Tartaglias method is based on the assumption that the solution to   x³ + 3cx + d = 0   had this form : x = p^1/3 + q^1/3

By cubing both sides and rearranging, one gets an expression where the constants can be identified:

x³ =p+3p^2/3 q^1/3 + 3p^1/3q ^2/3 + q  which leads to x³ =p+3(pq)^1/3 (p^1/3+q ^1/3 )+ q

x³ =p+3(pq)^1/3 x  + q and after rearranging :  x³ - 3(pq)^1/3 x - (p + q) = 0

If now the constants in this equation are compared with the original, it is easy to se that

c = (pq)^1/3 and   that     d = - (p+q)

which gives us:      q² + dq - c = 0      and       p² + dp - c = 0   which leads to:

q = p = (-d+ (d² + 4c³ )^1/2 )/2 . If p is chosen as the positive root and q the negative root we can, by inserting them in x = p^1/3 + q^1/3  get    x = p^1/3 +(-c³/p) ^1/3 which gives:

x=((-d+ (d² + 4c³ )^1/2 )/2)^1/3 - c/((-d+ (d² + 4c³ )^1/2 )/2)^1/3

x=(-d+ (d² + 4c³ )^1/6 )/2)^1/3 -2^1/3 c/(-d+ (d² + 4c³ )^1/6   

which is a root to the equation   x³ + 3cx + d = 0.

Tartaglia could now solve both  ax³ + bx² =c    and     ax³ +bx =c  and was also able, by the substitution   x = y- ^b/₃  to solve the general cubic equation    x³ + bx² +cx +d=0   by reducing it to the compressed form  ax³ +bx =c.

Here follows now the complete Ferro/Tartaglias method of solving the general cubic equation: x³ +bx² + cx +d = 0

Get rid of the x² -term by the substitution x = y- ^b/₃

(y- ^b/₃)³ +b(y- ^b/₃)² + c(y- ^b/₃) +d = 0 which leads to

(y³- 3y²b/3 +3yb²/9 – b³/27)+(by² - 2yb/3 + b³/9) +cy- ^bc/₃+d = 0

The underlined terms disappear and we get:

y³ +y(3b²/9 - 2b²/3 +c) - b³/27 + b³/9 - ^bc/₃ +d = 0

which equals to the reduced form:y³ +py + q =0

An example shows the rest of the solution:

x³ +9x = 6

say x = (u-v)

(u-v)³ +9(u-v) = 6 expand it :

u³-3u²v +3uv² –v³ +9u –9v = 6 rearrange:

u³–v³ – 3uv (u – v) +9(u –v) = 6 choose uv = 3 which gives

u³–v³ – 3*3 (u – v) +9(u –v) = 6 u³–v³ = 6

u³–v³ = 6 u³–v³ = 6 p–q = 6

uv = 3 u³v³ = 27 pq = 27 which leads to

an equation of second degree q² + 6q = 27 having the solution q = 3 which gives us p = 9 (only positive solutions were interesting) hence u³= 9 and v³ = 3

and x = u – v = ³9 – ³3

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