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Solving the cubic

Solving the cubic. 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.

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Solving the cubic

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  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 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.

  2. Let the cube be equal to the first power and constant and let DC and DF be two cubes the product of the sides of which, AB and BC, is equal to one-third the coefficient of x, and let the sum of these cubes be equal to the constant. I say that AC is the value of x. Now since AB x BC equals one-third the coefficient of x, 3(AB x BC) will equal the coefficient of x, and the product of AC and 3(AB x BC) is the whole first power, AC having been assumed to be x. But AC x 3(AB x BC) makes six bodies, three of which are AB x BC2and the other three, BC x AB2. Therefore these six bodies are equal to the whole first power, x3=cx+d Let DC, DF be cubes with sides AB and BC. s.t. (a) AB BC=1/3c. and (b) DC+DF=AB3+BC3=d. Then AC=x! Proof: AB BC=1/3c  3AB BC=c and under the assumption AC=x, AC(3ABxBC)=cx. Claim: AC(3ABxBC)=3ABxBC2+3BCxAB2. Notice that (AC=AB+BC) so that the equation is actually clear algebraically. By 7. AC(3ABxBC)=cx and by 8. AC(3ABxBC)=3ABxBC2+3BCxAB2 so 3ABxBC2+3BCxAB2=cx F E D A C B Girolamo Cardano (1501-1557)Ars Magna (The great art)

  3. and these [six bodies] plus the cubes DC and DF constitute the cube AE, according to the first proposition of Chapter VI. The cubes DC and DF are also equal to the given number. Therefore the cube AE is equal to the given first power and number, which was to be proved. It remains to be shown that 3AC(AB x BC) is equal to the six bodies. This is clear enough if I prove that AB(BC x AC) equals the two bodies ABxBC2 and BC x AB2, 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 — but this [i.e., AB x BE] is equal to the product of AB and (CD + DE); the product AB x DE is equal to the product CB x AB2, since all sides are equal to all sides; and therefore AC(AB x BC) is equal to AB x BC2 plus BC x AB2, as was proposed. 3ABxBC2+3BCxAB2+CD+DF=AE or (u+v)(3uv)=3u2v+3uv2+u3+v3=(u+v)3. DC+DF=d, this was assumption 3. So from 9,10,12 AE=cx+d and if AC=x then x3=cx+d. Remains to prove the claim 8. Enough to show AB(BC x AC) = AB x BC2+ BC x AB2 [by eliminating the common factor 3]. Since BE=ACxBC: AC(ABxBC)=ABxBE Since BE=CD+DE: ABxBE=AB(CD+DE) Since GE=AB & DG=CB: AB x DE= ABxGExDG=CB x AB2 So AC(ABxBC) = AB(CD+DE) = ABxBC2+ BC x AB2[since CD=BC2] F E D G A C B Girolamo Cardano (1501-1557)Ars Magna (The great art)

  4. Short version and the rule: Set x=u+v x3=(u+v)3=u3+3u2v+3uv2+v3. If (1) u3+v3=d and (2) 3uv=c, so 3u2v+3uv2=(u+v)3uv=cx. Then x3=cx+d. From (2) v=c/(3u). Substituted in (1) yields u3+(c/(3u))3=d  u6-du3+(c/3)3=0 Set U=u3, then U2-dU+(c/3)3=0 and Let V=v3, then V=d-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 The Great Art

  5. u3 u+v v u uv(u+v) v3 Binomial Theorem for Exponent 3 in Terms of Solids • (u+v)3=u3+3u2v+3uv2+v3=u3+3uv(u+v)+v3

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

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