已知1/a+1/b+1/c=1/(a+b+c),求证1/a3+1/b3+1/c3=1/(a3+b3+c3)神人们

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已知1/a+1/b+1/c=1/(a+b+c),求证1/a3+1/b3+1/c3=1/(a3+b3+c3)神人们

已知1/a+1/b+1/c=1/(a+b+c),求证1/a3+1/b3+1/c3=1/(a3+b3+c3)神人们
已知1/a+1/b+1/c=1/(a+b+c),求证1/a3+1/b3+1/c3=1/(a3+b3+c3)
神人们

已知1/a+1/b+1/c=1/(a+b+c),求证1/a3+1/b3+1/c3=1/(a3+b3+c3)神人们
等式1/x+1/y+1/z = 1/(x+y+z)等价于(x+y+z)(xy+yz+zx) = xyz,
等价于x²y+xy²+y²z+yz²+z²x+zx²+2xyz = 0,等价于(x+y)(y+z)(z+x) = 0.
因此由1/a+1/b+1/c = 1/(a+b+c)可得(a+b)(b+c)(c+a) = 0.
于是(a³+b³)(b³+c³)(c³+a³) = (a+b)(b+c)(c+a)(a²-ab+b²)(b²-bc+c²)(c²-ca+a²) = 0.
进而有1/a³+1/b³+1/c³ = 1/(a³+b³+c³).