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已知a+1/b=b+1/c=c+1/a,求证：(abc)^2=1
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已知a+1/b=b+1/c=c+1/a ,求证：(abc)^2=1
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答案和解析
a + 1/b = b + 1/c = c + 1/a,三项都乘上abc
a²bc + ac = ab²c + ab = abc² + bc
由a²bc + ac = ab²c + ab得abc(a - b) = a(b - c)
由ab²c + ab = abc² + bc得abc(b - c) = b(c - a)
由a²bc + ac = abc² + bc得abc(c - a) = c(a - b)
上面三项相乘得：
(abc)³(a - b)(b - c)(c - a) = abc(b - c)(c - a)(a - b),a ≠ b ≠ c
(abc)³ = abc,abc ≠ 0
(abc)² = 1
a²bc + ac = ab²c + ab = abc² + bc
由a²bc + ac = ab²c + ab得abc(a - b) = a(b - c)
由ab²c + ab = abc² + bc得abc(b - c) = b(c - a)
由a²bc + ac = abc² + bc得abc(c - a) = c(a - b)
上面三项相乘得：
(abc)³(a - b)(b - c)(c - a) = abc(b - c)(c - a)(a - b),a ≠ b ≠ c
(abc)³ = abc,abc ≠ 0
(abc)² = 1
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