$\frac{x+y}{z}$ + $\frac{y+z}{x}$ + $\frac{z+x}{y}$ - $\frac{x³+y³+z³}{xyz}$ = 2
⇔ $\frac{xy(x+y)}{xyz}$ + $\frac{yz(y+z)}{xyz}$ + $\frac{xz(z+x)}{xyz}$ - $\frac{x³+y³+z³}{xyz}$ = 2
⇔ $\frac{xy(x+y) + yz(y+z) + xz(z+x)- (x³+y³+z³)}{xyz}$ = 2
⇔ $\frac{(x²y+xy²) + (y²z +yz²) + (xz²+x²z) - (x³+y³+z³)}{xyz}$ = 2
⇔ $\frac{x²y+xy² + y²z +yz² + xz²+x²z - x³-y³-z³}{xyz}$ =2