66th Belarusian Mathematical Olympiad

Let $xyz=a$. Then $$\begin{array}{rl}(xy-z^2)(yz-x^2)(zx-y^2)& =\\ & =\left(\frac{a}{z}-z^2\right)\left(\frac{a}{x}-x^2\right)\left(\frac{a}{y}-y^2\right)=\frac{1}{a}(a-z^3)(a-x^3)(a-y^3)=\\ & =\frac{1}{a}(a^3-(x^3+y^3+z^3)a^2+(x^3{y}^3+y^3{z}^3+z^3{x}^3)a-x^3{y}^3{z}^3)=\\ & =[x^3{y}^3{z}^3=a^3]=(x^3{y}^3+y^3{z}^3+z^3{x}^3)-(x^3+y^3+z^3)a=A-Ba,(1)\end{array}$$ where $$A=x^3{y}^3+y^3{z}^3+z^3{x}^3\text{and}B=x^3+y^3+z^3.$$ Let $x{y}^2+x^{2}y+y{z}^2+y^{2}z+z{x}^2+x{z}^2=C$. Find the values of $A$, $B$, $C$. First, $$x^2+y^2+z^2=(x+y+z{)}^2-2(xy+yz+zx)=1-2\cdot (-1)=3.$$ Secondly, $$B=x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2)-C=(-1)\cdot 3-C=-3-C.(2)$$ Further, $$C=(x+y+z)(xy+yz+zx)-3xyz=-1\cdot (-1)-3a=1-3a.$$ From (2) it follows that $$B=-3-(1-3a)=3a-4.$$ Then $$x^2{y}^2+y^2{z}^2+z^2{x}^2=(xy+yz+zx{)}^2-2xyz(x+y+z)=(-1{)}^2-2\cdot (-1)=2a+1.$$ So, $$A=(xy+yz+zx)(x^2{y}^2+y^2{z}^2+z^2{x}^2)-Cxyz=(-1)(2a+1)-(1-3a)a=3a^2-3a-1.$$ Substituting $A$ and $B$ in (1), we obtain $$A-Ba=3a^2-3a-1-(3a-4)a=a-1=xyz-1,$$ as required.