HISTORY OF MATH HOMEWORK 5 SOLUTIONS 9.30, 32, 35, 37 ...

2 HISTORY OF MATH HOMEWORK 5 SOLUTIONS
9.35 Solve the system of equations
x 3 + 3xy 2 = x + y + 64
y 3 + 3yx 2 = x + y
Subtract the two equations and get (x − y) 3 = 64, so x − y = 4. Substitute y = x − 4 into the
second equation and get
y 3 + 6y 2 + 23
2 y − 1 = 0.
Let y = z − 2, to get
z 3 − 1z = 8. 2
So p = −1/6 and q = 4 and q 2 + p 3 = 16 − 1/216 = 3455/216. Cardano’s (actually Tartaglia’s
in this case!) formula gives
y = z − 2 = −2 +
and x = y + 4 as shown above.
[
4 +
√ ] 1/3 [
3455
+ 4 −
216
√ ] 1/3
3455
,
216
9.37 The equation x 3 + 3x = 36 has the root r = 3, but Cardano’s formula gives
Explain this using Bombelli’s methods.
r = [18 + √ 325] 1/3 + [18 − √ 325] 1/3 .
The numbers
α = [18 + √ 325] 1/3 , β = [18 − √ 325] 1/3
Satisfy
α + β = r, αβ = −p,
so they are the roots of the quadratic Y 2 − rY − p = 0, where r = 3 and p = 1. The quadratic
formula then gives
α, β = 1[3 ± √ 13].
2
You can check that
( 1 [3 + √ 13] ) 3 √
2 = 18 + 325,
( 1 [3 − √ 13] ) 3 √
2 = 18 − 325,
so
α = 1 2 [3 + √ 13], β = 1 2 [3 − √ 13].