# Solving the cubic

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```					              Solving the cubic
• Three people were involved in the solution of the cubic:
del Ferro (1465-1526), Niccolò Tartaglia (1499-1557)
and Girolamo Cardano (1501-1557). Del Ferro and
Tartaglia. There were thought to be two cases x3+cx=d
called the cosa and x3=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 x3+bx2=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.
Girolamo Cardano (1501-1557)
Ars Magna (The great art)
1. Let the cube be equal to the first        1.   x3=cx+d
power and constant                        2.   Let DC, DF be cubes with sides AB
2. and let DC and DF be two cubes                 and BC.
3. the product of the sides of which, AB     3. s.t. (a) AB BC=1/3c.
and BC, is equal to one-third the         4. and (b) DC+DF=AB3+BC3=d.
coefficient of x,                         5. Then AC=x!
4. and let the sum of these cubes be         Proof:
equal to the constant.
5. I say that AC is the value of x.          6. AB BC=1/3c  3AB BC=c
6. Now since AB x BC equals one-third        7. and under the assumption AC=x,
the coefficient of x, 3(AB x BC) will          AC(3ABxBC)=cx.
equal the coefficient of x,               8. Claim:
7. and the product of AC and 3(AB x BC)           AC(3ABxBC)=3ABxBC2+3BCxAB2.
is the whole first power, AC having            Notice that (AC=AB+BC) so that the
been assumed to be x.                          equation is actually clear
algebraically.
8. But AC x 3(AB x BC) makes six             9. By 7. AC(3ABxBC)=cx and by 8.
bodies, three of which are AB x BC2            AC(3ABxBC)=3ABxBC2+3BCxAB2 so
and the other three, BC x AB2.                 3ABxBC2+3BCxAB2=cx
9. Therefore these six bodies are equal
to the whole first power,             F            E

D

A             C
B
Girolamo Cardano (1501-1557)
Ars Magna (The great art)
10. and these [six bodies] plus the cubes      10.   3ABxBC2+3BCxAB2+CD+DF=AE or
DC and DF constitute the cube AE,                (u+v)(3uv)=3u2v+3uv2+u3+v3=(u+v)3.
according to the first proposition of      11.   DC+DF=d, this was assumption 3.
Chapter VI.                                12.   So from 9,10,12 AE=cx+d and if
11. The cubes DC and DF are also equal               AC=x then x3=cx+d.
to the given number.                       13.   Remains to prove the claim 8.
12. Therefore the cube AE is equal to the      14.   Enough to show
given first power and number, which              AB(BC x AC) = AB x BC2+ BC x AB2
was to be proved.                                [by eliminating the common factor
13. It remains to be shown that 3AC(AB x             3].
BC) is equal to the six bodies.            15.   Since BE=ACxBC:
14. This is clear enough if I prove that                   AC(ABxBC)= ABxBE
AB(BC x AC) equals the two bodies          16.   Since BE=CD+DE:
ABxBC2 and BC x AB2,                                   ABxBE=AB(CD+DE)
15. for the product of AC and (AB x BC) is     17.   Since GE=AB & DG=CB:
equal to the product of AB and the                     AB x DE= ABxGExDG=CB x
surface BE — since all sides are equal           AB  2
to all sides —
16. but this [i.e., AB x BE] is equal to the   18.   So AC(ABxBC) = AB(CD+DE) =
product of AB and (CD + DE);                     ABxBC2+ BC x AB2 [since CD=BC2]
F                   E
17. the product AB x DE is equal to the
product CB x AB2, since all sides are
equal to all sides;                                         D       G
18. and therefore AC(AB x BC) is equal to
AB x BC2 plus BC x AB2, as was
proposed.                                        A                  C
B
The Great Art
Short version and the rule:                    As Cardano says:
1. Set x=u+v                                   The rule, therefore, is:
2. x3=(u+v)3=u3+3u2v+3uv2+v3.                  When the cube of one-third
3. If (1) u3+v3=d and (2) 3uv=c, so            the coefficient of x is not
3u2v+3uv2=(u+v)3uv=cx.                     greater than the square of
4. Then x3=cx+d.                               one-half the constant of the
equation, subtract the former
5. From (2) v=c/(3u).                          from the latter and add the
6. Substituted in (1) yields                   square root of the remainder
u3+(c/(3u))3=d  u6-du3+(c/3)3=0           to one-half the constant of
7. Set U=u3, then U2-dU+(c/3)3=0 and           the equation and, again,
subtract it from the same
U  d  ( d ) 2  ( c )3
2     2         3
half, and you will have, as
was said, a binomium and its
8.   Let V=v3, then V=d-U, so                  apotome, the sum of the
cube roots of which
V  d  ( d ) 2  ( c )3
2     2         3                 constitutes the value of x
9.   Finally

x 3 U 3 V 3        d
2    ( d ) 2  ( c )3  3
2         3
d
2    ( d ) 2  ( c )3
2         3
Binomial Theorem for Exponent 3
in Terms of Solids
• (u+v)3=u3+3u2v+3uv2+v3=u3+3uv(u+v)+v3

u3

v   u     v3   uv(u+v)
Examples
• x3=6x+40                              • Problems with x3=15x+4,
– c=6, (c/3)3=23=8                    the formula yields
– d=40, (d/2)2=202=400                  – c=15, (c/3)3=53=125
– 400-8=392 and                         – d=4, (d/2)2=22=4
– 125-4=121 and se
x  3 20  392  3 20  392
x  3 2   121  3 2   121
• x3=6x+6                                      which contains complex
numbers. But actually there
– c=6, (c/3)3=23=8
are three real solutions:
– d=6, (d/2)2=32=9
4,-2+ √3 and -2-√3.
– 9-8=1 and thus
x  3 3 1  3 3 1  3 4  3 2        • Also what about the other
possible solutions?
Rafaeleo Bombelli (1526-1572) started to calculate cube roots of complex numbers

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