solve a equation

(a - b)(a^2 + ab + b^2) = a^3 - b^3
(a - b)(a^2 - 2ab + b^2 + 3ab) = a^3 - b^3
(a - b)[(a - b)^2 + 3ab] = a^3 - b^3
(a - b)^3 + 3ab(a - b) = a^3 - b^3
Compare with x^3 + x = 15 :
x = (a - b)
1 = 3ab................. (1)
15 = a^3 - b^3........(2)
(1)^3 :
1 = 27(a^3)(b^3)
(a^3)(b^3) = 1/27.......(3)
(2)^2 :
a^6 + b^6 - 2(a^3)(b^3) = 225....(4)
(4) + (3) * 4 :
a^6 + b^6 + 2(a^3)(b^3) = 225 + 4/27
(a^3 + b^3)^2 = 6079/27
a^3 + b^3 = 15.00493746.....(5)
(5) + (2) :
2a^3 = 15 + 15.00493746
a^3 = 15.00246873 , sub it to (2):
b^3 = 15.00246873 - 15 = 0.00246873
a = 2.466347365 , b = 0.135152639
x = (a - b) = 2.466347365 - 0.135152639 = 2.331194726......
So one root is 2.331194726 , and you can easy to find other two roots
by the quadratic equation which is given by
(x^3 + x - 15) / (x - 2.331194726...) = 0





2009-09-29 19:34:14 補充：
The quadratic equation is :

x^2 + 2.331194726x + 6.434468851 = 0

△ = 2.331194726 ^ 2 - 4(1)(6.434468851) < 0

So the quadratic equation is no real roots.

So x = 2.331194726.... is the only root for x is real number.

Cubic equation is out of syllabus for secondary school.

2009-09-29 19:36:44 補充：
The quadratic equation is :

x^2 + 2.331194726x + 6.434468851 = 0

△ = 2.331194726 ^ 2 - 4(1)(6.434468851) < 0

So the quadratic equation have no real roots.

So x = 2.331194726.... is the only root for x is a real number.

Cubic equation is out of syllabus for secondary school.