標題:
pure maths prob~
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發問:
Find k s.t. x^3+y^3+z^3 +xyz is divisible by (x+y+z)
最佳解答:
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)
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