Transcript Slides for Lecture 22

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
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
F
to the whole first power,
x3=cx+d
Let DC, DF be cubes with sides AB
and BC.
3. s.t. (a) AB BC=1/3c.
4. and (b) DC+DF=AB3+BC3=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)=3ABxBC2+3BCxAB2.
Notice that (AC=AB+BC) so that the
equation is actually clear
algebraically.
9. By 7. AC(3ABxBC)=cx and by 8.
AC(3ABxBC)=3ABxBC2+3BCxAB2 so
3ABxBC2+3BCxAB2=cx
1.
2.
E
D
A
B
C
Girolamo Cardano (1501-1557)
Ars Magna (The great art)
10. and these [six bodies] plus the cubes
DC and DF constitute the cube AE,
according to the first proposition of
Chapter VI.
11. The cubes DC and DF are also equal
to the given number.
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 3AC(AB x
BC) is equal to the six bodies.
14. This is clear enough if I prove that
AB(BC x AC) equals the two bodies
ABxBC2 and BC x AB2,
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. the product AB x DE is equal to the
product CB x AB2, since all sides are
equal to all sides;
18. and therefore AC(AB x BC) is equal to
AB x BC2 plus BC x AB2, as was
proposed.
10.
11.
12.
13.
14.
15.
16.
17.
18.
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
2
AB
So AC(ABxBC) = AB(CD+DE) =
2+ BC x AB2 [since CD=BC2]
ABxBC
F
E
D
A
G
B
C
The Great Art
Short version and the rule:
1. Set x=u+v
2. x3=(u+v)3=u3+3u2v+3uv2+v3.
3. If (1) u3+v3=d and (2) 3uv=c, so
3u2v+3uv2=(u+v)3uv=cx.
4. Then x3=cx+d.
5. From (2) v=c/(3u).
6. Substituted in (1) yields
u3+(c/(3u))3=d  u6-du3+(c/3)3=0
7. Set U=u3, then U2-dU+(c/3)3=0 and
U 
d
2

( d2 )  ( c3 )
2
8.
Let V=v3, then V=d-U, so
9.
V 
Finally
x 
3
d
2
U 

3
( d2 )  ( c3 )
2
V 
3
d
2

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
( d2 )  ( c3 ) 
2
3
3
d
2

( d2 )  ( c3 )
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,
the formula yields
– c=6, (c/3)3=23=8
– d=40, (d/2)2=202=400
– 400-8=392 and
x
3
20 
392 
3
20 
392
– c=15, (c/3)3=53=125
– d=4, (d/2)2=22=4
– 125-4=121 and se
x
• x3=6x+6
3
31 
3
31 
3
4
2
 121 
3
2
 121
which contains complex
numbers. But actually there
are three real solutions:
4,-2+ √3 and -2-√3.
– c=6, (c/3)3=23=8
– d=6, (d/2)2=32=9
– 9-8=1 and thus
x
3
3
2
• Also what about the other
possible solutions?
Rafaeleo Bombelli (1526-1572) started to calculate cube roots of complex numbers