Solving the cubic

1
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 x3cxd called the cosa and x3cxd 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 x3bx2d, 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
  Ferros solution.

2
Girolamo Cardano (1501-1557)Ars Magna (The great
art)

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 BC2 and the other three, BC x AB2.
   
9. Therefore these six bodies are equal to the whole
   first power,

• x3cxd
• Let DC, DF be cubes with sides AB and BC.
• s.t. (a) AB BC1/3c.
• and (b) DCDFAB3BC3d.
• Then ACx!
• Proof
• AB BC1/3c ? 3AB BCc
• and under the assumption ACx, AC(3ABxBC)cx.
• Claim AC(3ABxBC)3ABxBC23BCxAB2. Notice that
  (ACABBC) so that the equation is actually clear
  algebraically.
• By 7. AC(3ABxBC)cx and by 8. AC(3ABxBC)3ABxBC23
  BCxAB2 so 3ABxBC23BCxAB2cx

3
Girolamo Cardano (1501-1557)Ars Magna (The great
art)

1. and these six bodies plus the cubes DC and DF
   constitute the cube AE, according to the first
   proposition of Chapter VI.
2. The cubes DC and DF are also equal to the given
   number.
3. Therefore the cube AE is equal to the given first
   power and number, which was to be proved.
4. It remains to be shown that 3AC(AB x BC) is equal
   to the six bodies.
5. This is clear enough if I prove that AB(BC x AC)
   equals the two bodies ABxBC2 and BC x AB2,
6. 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
7. but this i.e., AB x BE is equal to the product
   of AB and (CD DE)
8. the product AB x DE is equal to the product CB x
   AB2, since all sides are equal to all sides
9. and therefore AC(AB x BC) is equal to AB x BC2
   plus BC x AB2, as was proposed.

1. 3ABxBC23BCxAB2CDDFAE or (uv)(3uv)3u2v3uv2u
   3v3(uv)3.
2. DCDFd, this was assumption 3.
3. So from 9,10,12 AEcxd and if ACx then x3cxd.
4. Remains to prove the claim 8.
5. Enough to show AB(BC x
   AC) AB x BC2 BC x AB2 by eliminating the
   common factor 3.
6. Since BEACxBC AC(ABxBC) ABxBE
7. Since BECDDE ABxBEAB(CDDE)
8. Since GEAB DGCB AB x DE ABxGExDGCB x AB2
9. So AC(ABxBC) AB(CDDE) ABxBC2 BC x AB2
   since CDBC2

4
The Great Art

• Short version and the rule
• Set xuv
• x3(uv)3u33u2v3uv2v3.
• If (1) u3v3d and (2) 3uvc, so
  3u2v3uv2(uv)3uvcx.
• Then x3cxd.
• From (2) vc/(3u).
• Substituted in (1) yields u3(c/(3u))3d ?
  u6-du3(c/3)30
• Set Uu3, then U2-dU(c/3)30 and
• Let Vv3, then Vd-U, so
• Finally

• 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

5
Binomial Theorem for Exponent 3 in Terms of Solids

• (uv)3u33u2v3uv2v3u33uv(uv)v3

6
Examples

• x36x40
• c6, (c/3)3238
• d40, (d/2)2202400
• 400-8392 and
• x36x6
• c6, (c/3)3238
• d6, (d/2)2329
• 9-81 and thus

• Problems with x315x4, the formula yields
• c15, (c/3)353125
• d4, (d/2)2224
• 125-4121 and se
• which contains complex numbers. But actually
  there are three real solutions
• 4,-2 v3 and -2-v3.
• Also what about the other possible solutions?

Rafaeleo Bombelli (1526-1572) started to
calculate cube roots of complex numbers