pure maths prob~

where is k?

2009-09-05 11:56:00 補充：
Find k s.t. x3 + y3 + z3 + kxyz is divisible by (x+y+z)
(x + y + z)3
= x3 + y3 + z3 + 3x2y + 3x2z + 3xy2 + 3y2z + 3xz2 + 3yz2 + 6xyz
= x3 + y3 + z3 + 3x2y + 3x2z + 3x3 - 3x3 + 3xy2 + 3y2z +3y3 - 3y3 + 3xz2 + 3yz2 + 3z3 - 3z3 + 6xyz
= x3 + y3 + z3 + 3x2(x + y + z) - 3x3 + 3y2(x + y + z) - 3y3 + 3z2(x + y + z) - 3z3 + 6xyz
= -2(x3 + y3 + z3) + 3(x2 + y2 + z2)(x + y + z) + 6xyz
(x3 + y3 + z3) - 3xyz = (3/2)(x2 + y2 + z2)(x + y + z) - (x + y + z)3/2
(x3 + y3 + z3) - 3xyz = (x + y + z)[(3/2)(x2 + y2 + z2) - (x + y + z)2/2]
Hence k = -3

2009-09-05 12:01:03 補充：
Alternately, consider f(x) = x^3 + y^3 + z^3 + kxyz is divisible by [x + (y + z)]
f(- y - x) = (-y-z)^3 + y3 + z3 + k(-y-z)yz = 0
-3y^2z - 3yz^2 - ky^2z - kyz^2 = 0
-(3 + k)(y^2z + yz^2) = 0 this holds true for all values of y and z => k = -3

2009-09-05 12:01:36 補充：
it should be f(-y - z)