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).