數學知識交流---代數間題

x + y = a (Given)
x^2 + y^2 = b (Given)

(a) x + y = a
Square on both sides
(x + y)^2 = a^2
x^2 + 2xy + y^2 = a^2
b + 2xy = a^2
-2xy = b – a^2
-xy = (b – a^2)/2 ------------- (1)

x^3 + y^3 = (x + y) (x^2 – xy + y^2)
= a (b – xy)
= a(b +(b – a^2)/2)
= ab +a(b-a^2)/2
= (2ab +ab – a^3)/2
= (3ab – a^3)/2

x^3 + y^3 = (3ab – a^3)/2

(b) x^4 + y^4 = (x^2 +(2^0.5)xy + y^2) (x^2 - (2^0.5)xy + y^2)
= (b +(2^0.5)xy ) (b - (2^0.5)xy)
= b^2 -2(x^2)(y^2)
= b^2 -2(xy)^2 ----------------------- (2)

From equation (1) above,
-xy = (b – a^2)/2
xy = (a^2 – b)/2
(xy)^2 = (a^2 – b)^2/4 ------------ (3)

Substitute equation (3) into equation (2)
x^4 + y^4 = b^2 –[2(a^2 – b)^2]/4
x^4 + y^4 = b^2 – [(a^2 – b)^2]/2
x^4 + y^4 = b^2 – (1/2)[(a^2 – b)^2]

x^4 + y^4 = b^2 – (1/2)[(a^2 – b)^2]

Check:
Let x = 1, y = 2
a = x + y = 1+2 = 3

x^2 + y^2 = b
1^2 + 2^2 = b
b = 5

x^4 + y^4 = 1^4 + 2^4 = 17
b^2 – 1/2 [(a^2 – b)^2] = 5^2 – 1/2 [(3^2 – 5)^2]
= 25 – 1/2 [4^2]
= 25 – 8
= 17

2011-07-05 11:11:20 補充：
You have to know the factorization of
x^4 + y^4 = [x^2 + (2^0.5)xy + y^2] [x^2 - (2^0.5)xy + y^2]
where (2^0.5) is square root of 2

x^3 + y^3=x^2 + y^2+x + y
x^3 + y^3=a+b

x^4 + y^4=x^2 + y^2+x^2 + y^2
x^4 + y^4=b*b
x^4 + y^4=b^2

x + y = a

x^2 + y^2 = b
(x^2 + 2xy + y^2) - 2xy = b
(x + y)^2 - 2xy = b
a^2 - 2xy = b
2xy = a^2 - b
xy = (a^2 - b)/2

(a) x^3 + y^3
= (x^3 + 3x^2 y + 3xy^2 + y^3) - (3x^2 y + 3xy^2)
= (x + y)^3 - 3xy (x + y)
= a^3 - 3 [(a^2 - b)/2] (a)
= a^3 - 3a(a^2 - b)/2
= [2a^3 - 3a(a^2 - b)] / 2
= [2a^3 - 3a^3 + 3ab] / 2
= (3ab - a^3)/2

(b) x^4 + y^4
= (x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + y^4) - (4x^3 y + 6x^2 y^2 + 4xy^3)
= (x + y)^4 - (4x^3 y + 6x^2 y^2 + 4xy^3)
= (x + y)^4 - [4xy (x^2 + y^2) + 6(xy)^2]
= a^4 - {4 * (3ab - a^3)/2 * b + 6 [(3ab - a^3)/2]^2}
= a^4 - [2b (3ab - a^3) + 3/2 (3ab - a^3)^2]
= a^4 - [2b (3ab - a^3) + 3/2 (9a^2 b^2 - 6a^4 b + a^6)]
= a^4 - [(6ab^2 - 2a^3 b) + (27/2 a^2 b^2 - 9a^4 b + 3/2 a^6)]
= a^4 - (6ab^2 - 2a^3 b + 27/2 a^2 b^2 - 9a^4 b + 3/2 a^6)
= a^4 - 6ab^2 + 2a^3 b - 27/2 a^2 b^2 + 9a^4 b - 3/2 a^6
= - 3/2 a^6 + a^4 + 9a^4 b + 2a^3 b - 27/2 a^2 b^2 - 6ab^2

x^3 + y^3
樓上尾二個步計錯左
正確係
(2ab-a^3+ab)/2
(3ab-a^3)/2

x+y=a
b+2xy=a^2
xy=(a^2-b)/2
___________
x^3 + y^3
(x+y)(x^2-((a^2-b)/2)+y^2)
a(b-(a^2-b/2))
ab-(a^3-ab/2)
(2ab-a^3+ab)/2
(4ab-a^3)/2

2011-06-21 21:00:09 補充：
x^2 + y^2 = b
x^4+2x^2y^2+y^4=b^4
x^4+y^4=(b^4)-2x^2y^2
__________________
(b^4)-2x^2y^2
As (xy) equal to (a^2-b)/2
(b^4)-2((a^2-b)/2)^2
b^4-((a^2-b)^2/2)
((2b^4-a^2+2a^2b-b^2b)/2)

2011-06-21 23:07:37 補充：
Yes,something is wrong,thank you for your remind!

2011-06-22 17:05:44 補充：
Correction:
x^2 + y^2 = b
x^4+2x^2y^2+y^4=b^2
x^4+y^4=(b^2)-2x^2y^2
__________________
(b^2)-2x^2y^2
b^2-(a^2-b)^2/2
(2b^2-a^2+2a^2b+b^2)/2
(3b^2+2a^2b-a^2)/2