Re: [中學] 今日鳳中教甄

[motivic (Ian)]

※ 引述《shingai (吸收正能量)》之銘言：
: #2 設a, b, c 為x^3-x^2+2x-3=0 之三根
:    求
:    a^3/[(a^2-b^2)(a^2-c^2)]+ b^3/[(b^2-a^2)(b^2-c^2)]+ c^3/[(c^2-b^2)(c^2-a^2)]
:    之值


Let S=a^3/[(a^2-b^2)(a^2-c^2)]+ b^3/[(b^2-a^2)(b^2-c^2)]+
                                                  c^3/[(c^2-b^2)(c^2-a^2)]

                         a^3(b+c)     b^3(c+a)     c^3(a+b)
Then S*(a+b)(b+c)(c+a)= ----------- + ---------- + --------
                        (a-b)(a-c)    (b-c)(b-a)   (c-a)(c-b)

        a^3(x-b)(x-c)         b^3(x-c)(x-z)         c^3(x-a)(x-b)
H(x):= ------------------- + ------------------- + ---------------
        (a-b)(a-c)            (b-c)(b-a)            (c-a)(c-b)

Hence H(x)=x^3 for x=a,b,c.

Since deg(H)<=2, we have  H(x)=x^3 -  (x-a)(x-b)(x-c).

Note that the coefficient of x in H(x) is the same as -S*(a+b)(b+c)(c+a)

Thus, S=(ab+bc+ca)/[(a+b)(b+c)(c+a)]=-2.


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※ 編輯: motivic (114.45.201.100), 05/03/2015 21:48:59
推 shingai     : 了解！ 感謝