根與係數關係變化(二)

x^4+2x^3+x+1=0 => -x^3 = (x+1)/(x+2) = 1- 1/(x+2)
=> -(a^3 + b^3 + c^3 + d^3)= 4 - [1/(a+2) + 1/(b+2) + 1/(c+2) + 1/(d+2)]
令 f(x) = x^4+2x^3+x+1 = (x-a)(x-b)(x-c)(x-d)
則 f'(x) / f(x) = 1/(x-a) + 1/(x-b) + 1/(x-c) + 1/(x-d)
=> 7 = f'(-2) / f(-2) = 1/(-2-a) + 1/(-2-b) + 1/(-2-c) + 1/(-2-d)
=> a^3 + b^3 + c^3 + d^3 = -11 = s^3
原式 = (s^3-d^3)(s^3-c^3)(s^3-b^3)(s^3-a^3)
= f(s)f(sw)f(sw^2) ( w為x^3=1的單位虛根，詳見註1)
= (-10s-21)(-10sw-21)(-10sw^2 - 21) = (-10s)^3 - 21^3 = 11000-9261
= 1739

註1：a^3 - b^3 = (a-b)(aw-b)(aw^2-b) = (a-b)(a-bw)(a-bw^2)



2009-02-20 19:37:36 補充：
a^3 + b^3 + c^3 + d^3 也可以用別的方法算

x^3 = -2x^2 - 1 - 1/x

因此 a^3 + b^3 + c^3 + d^3 = -8 - 4 +1 = -11