若x3+px+q能被(x+a)2整除, 證明(p/3)3+(q/2)2=0

Since there is no remainder, x^3 + px + q = (x+a)^2Q(x), where Q(x) is the quotient .Put x = -a, we get -a^3 -ap + q = 0....................(1)
3x^2 + p = 2(x+a)Q(x) + (x+a)^2[dQ(x)/dx]. Put x = -a, we get
3a^2 + p = 0 ...................(2)
(1) x 3 we get -3a^3 -3ap + 3q = 0..............(3)
(2) x a we get 3a^3 +ap = 0..............(4)
(3) + (4) we get -2ap + 3q = 0, or a = 3q/2p.
Substitute this into (2), we get
3(3q/2p)^2 + p = 0
27q^2/4p^2 + p = 0
27q^2 + 4p^3 = 0
q^2/4 + p^3/27 = 0
(q/2)^ + (p/3)^3 = 0.

2008-08-04 16:01:20 補充：
Differentiate both sides of line 1 we get line 3.

2008-08-05 10:24:02 補充：
If you do not know calculus, you can put x^3 + px + q = (x+a)^2(x+b). Then set up 3 equations in
p,q,a and b, then eliminate a and b to get to the result.