• Three people were involved in the solution of the cubic: del Ferro (1465-1526), Niccol ò Tartaglia (1499-1557) and Girolamo Cardano (1501-1557). Del Ferro and
Tartaglia. There were thought to be two cases x 3 +cx=d called the cosa and x 3 =cx+d with positive c and d.
• Del Ferro solved the cosa and kept his result secret.
• Tartaglia then first claimed to be able to solve equations of the kind x 3 +bx 2 =d, but then also solved all cubic equations.
• Cardano was given the formula by Tartaglia and found and published proofs for the formulas for all cubic equations. He was also aware of del Ferro’s solution.

1. Let the cube be equal to the first power and constant
2. and let DC and DF be two cubes
3. the product of the sides of which, AB and BC , is equal to one-third the coefficient of x ,
4. and let the sum of these cubes be equal to the constant.
5. I say that AC is the value of x .
6. Now since AB x BC equals one-third the coefficient of x , 3( AB x BC ) will equal the coefficient of x ,
7. and the product of AC and 3( AB x BC ) is the whole first power, AC having been assumed to be x .
8. But AC x 3( AB x BC ) makes six bodies, three of which are AB x BC 2 and the other three, BC x AB 2 .
9. Therefore these six bodies are equal to the whole first power,
F
1.
x 3 =cx+d
2.
Let DC, DF be cubes with sides AB and BC.
3.
s.t. (a) AB BC=1/3c.
4.
and (b) DC+DF=AB 3 +BC 3 =d.
5.
Then AC=x!
Proof:
6.
AB BC=1/3c  3AB BC=c
7.
and under the assumption AC=x,
AC(3ABxBC)=cx .
8.
Claim:
AC(3ABxBC)=3ABxBC 2 +3BCxAB 2 .
Notice that (AC=AB+BC) so that the equation is actually clear algebraically.
9.
By 7. AC(3ABxBC)=cx and by 8.
AC(3ABxBC)=3ABxBC 2 +3BCxAB
3ABxBC 2 +3BCxAB 2 =cx
2 so
E
A
D
B
C

10. and these [six bodies] plus the cubes
DC and DF constitute the cube AE , according to the first proposition of
Chapter VI.
10.
3ABxBC 2 +3BCxAB
(u+v)(3uv)=3u 2
2 +CD+DF=AE or v+3uv 2 +u 3 +v 3 =(u+v) 3 .
11.
DC+DF=d , this was assumption 3.
11. The cubes DC and DF are also equal to the given number.
12.
So from 9,10,12 AE=cx+d and if
AC=x then x 3 =cx+d .
13.
Remains to prove the claim 8.
12. Therefore the cube AE is equal to the given first power and number, which was to be proved.
13. It remains to be shown that 3 AC ( AB x
BC ) is equal to the six bodies.
14.
Enough to show
AB(BC x AC) = AB x BC 2 + BC x AB 2
[by eliminating the common factor
3].
14. This is clear enough if I prove that
AB ( BC x AC ) equals the two bodies
ABxBC 2 and BC x AB 2 ,
15.
Since BE=ACxBC:
AC(ABxBC)= ABxBE
16.
Since BE=CD+DE:
ABxBE= AB(CD+DE)
15. for the product of AC and ( AB x BC ) is equal to the product of AB and the surface BE — since all sides are equal to all sides —
16. but this [i.e., AB x BE ] is equal to the product of AB and ( CD + DE );
17.
Since GE=AB & DG=CB:
AB x DE= ABxGExDG=CB x
AB 2
18.
So AC(ABxBC) = AB(CD+DE) =
ABxBC 2 + BC x AB 2 [since CD=BC 2 ]
17. the product AB x DE is equal to the product CB x AB 2 , since all sides are equal to all sides;
18. and therefore AC ( AB x BC ) is equal to
AB x BC 2 plus BC x AB 2 , as was proposed.
A
D
B
G
C

Short version and the rule:
1.
Set x=u+v
2.
x 3 =(u+v) 3 = u 3 + 3u 2 v+3uv 2 + v 3 .
3.
If (1) u 3 +v 3 =d and (2) 3uv=c , so
3u 2 v+3uv 2 =(u+v)3uv= cx .
4.
Then x 3 =cx+d.
5.
From (2) v=c/(3u).
6.
Substituted in (1) yields u 3 +(c/(3u)) 3 =d  u 6 -du 3 +(c/3) 3 =0
7.
Set U=u 3 , then U 2 -dU+(c/3) 3 =0 and
U
 d
2

( d
2
)
2 
( c
3
)
3
8.
Let V=v 3 , then V=d-U, so
V
9.
Finally
 d
2
 x

3 U

3 V
( d
2
)
2 
( c
3
)
3

3 d
2

( d
2
)
2 
( c
3
)
3 
3
As Cardano says:
The rule, therefore, is:
When the cube of one-third the coefficient of x is not greater than the square of one-half the constant of the equation, subtract the former from the latter and add the square root of the remainder to one-half the constant of the equation and, again, subtract it from the same half, and you will have, as was said, a binomium and its apotome , the sum of the cube roots of which constitutes the value of x d
2

( d
2
)
2 
( c
3
)
3

3
3
2
2
3
3
3 u 3 v u v 3 uv(u+v)

• x 3 =6x+40
– c=6, (c/3) 3 =2 3 =8
– d=40, (d/2) 2 =20 2 =400
– 400-8=392 and x

3 20

392

3 20

• x 3 =6x+6
– c=6, (c/3) 3 =2 3 =8
– d=6, (d/2) 2 =3 2 =9 x

– 9-8=1 and thus
3 3

1

3 3

1

3 4

3
392
2
• Problems with x 3 =15x+4, the formula yields
– c=15, (c/3) 3 =5 3 =125
– d=4, (d/2) 2 =2 2 =4
– 125-4=121 and se x

3
2
 
121

3
2
 
121 which contains complex numbers. But actually there are three real solutions:
4,-2+ √3 and -2-√3.
• Also what about the other possible solutions?
Rafaeleo Bombelli (1526-1572) started to calculate cube roots of complex numbers