二項式問題

(1)
If x+y = 3, y+z = 2
Then (y+z)-(x+y) = 2 - 3
z - x = -1

x^2 + y^2 + z^2 + xy + yz - zx
= 1/2 (2x^2 + 2y^2 + 2z^2 + 2xy + 2yz - 2zx)
= 1/2 (x^2 + 2xy + y^2 + z^2 - 2zx + x^2 + y^2 + 2yz + z^2)
= 1/2 [ (x+y)^2 + (z-x)^2 + (y+z)^2 ]
= 1/2 [ 3^2 + (-1)^2 + 2^2 ]
= 1/2 (9+1+4)
= 7

(2)
x^2 + y^2 + z^2 - xy - yz - zx = 19
2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx = 38
(x^2 - 2xy + y^2) + (z^2 - 2zx + x^2) + (y^2 - 2yz + z^2) = 38
(x-y)^2 + (z-x)^2 + (y-z)^2 = 38
2^2 + (z-x)^2 + (y-z)^2 = 38
(z-x)^2 + (y-z)^2 = 34 ................(*)

Let y-z = M
Then z-x = -(y-z)-(x-y) = -M-2
Put into (*)
(-M-2)^2 + M^2 = 34
M^2 + 4M + 4 + M^2 = 34
M^2 + 2M + 2 = 17
用完成平方法 (By completing the square)
(M + 1)^2 + 1 = 17
(M + 1)^2 = 16
(M + 1) = 4 or -4
M = 3 or -5
因為 M = y - z
所以，y - z 的最大值為 3