51. Let $a,b,c$ is nonzero real numbers such that $a+b+c\not=0$ and $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$
Prove that for all odd integers $n$ $$\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$$[/quote]
solution:คูณสมการที่ให้มาทั้งสองฝั่งด้วย$abc(a+b+c)$จะได้$$ab(a+b+c)+bc(a+b+c)+ca(a+b+c)=abc$$ $$3abc+a^2b+b^2a+b^2c+c^2b+c^2a+a^2c=abc$$ $$abc+a^2b+b^2a+b^2c+c^2b+c^2a+a^2c+abc=0$$ $$(a+b)(b+c)(c+a)=0$$ จากสมการที่ได้ symmetric โดยไม่เสียนัยสำคัญให้$a+b=0$ ได้ $a=-b$ จะได้ $a^n=-b^n$ $$\frac{1}{c^n}=\frac{1}{c^n}$$ $$\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$$ $$\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$$
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