In the year 1530 Fiore had challenged a mathematician from Brescia, Tonioni da Coi into a, for  that time, 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 another kind namely:    ax³ +bx =c.

When in February 1535 Fiore did challenge Tartaglia he naturally gave him problems to solve that of course were of the kind he managed himself.

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 reduce   the general cubic equation
         x³ + bx² +cx +d=0   to    the compressed form  ax³ +bx =c.

Here now follows 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 (blue) terms disappear and with them y² - term, 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,
where      p = 3b²/9 - 2b²/3 +c
and         q = - b³/27 + b³/9 - ^bc/₃ +d

An example now shows the rest of the solution:

x³ + 9x = 6     say x = (u - v)

(u - v)³ +9(u - v) = 6          expand

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    and  p – q = 6

uv = 3 u³v³ = 27  and  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 finally x = u – v = ³9 – ³3