1. **Problem Statement:** Given $$L= \frac{xy + zx - x^2}{(\log_g k)}, \quad M= \frac{yz + xy - y^2}{(\log_g k)}, \quad N= \frac{zx + yz - z^2}{(\log_g k)}$$ and the condition $$L = M = N,$$ prove that $$x^y y^x = y^z z^y = z^x x^z.$$ 2. **Understanding the problem:** We have three expressions equal to each other. Since the denominators are the same, the numerators must be equal: $$xy + zx - x^2 = yz + xy - y^2 = zx + yz - z^2.$$ 3. **Set the numerators equal:** From $$xy + zx - x^2 = yz + xy - y^2,$$ subtract $$xy$$ from both sides: $$zx - x^2 = yz - y^2.$$ Rewrite as: $$z x - x^2 = y z - y^2.$$ Similarly, from $$yz + xy - y^2 = zx + yz - z^2,$$ subtract $$yz$$ from both sides: $$xy - y^2 = zx - z^2.$$ 4. **Rearranging the equations:** Equation 1: $$z x - x^2 = y z - y^2 \implies z x - y z = x^2 - y^2 \implies z(x - y) = (x - y)(x + y).$$ If $$x \neq y,$$ divide both sides by $$x - y$$: $$z = x + y.$$ Equation 2: $$xy - y^2 = zx - z^2 \implies xy - zx = y^2 - z^2 \implies x(y - z) = (y - z)(y + z).$$ If $$y \neq z,$$ divide both sides by $$y - z$$: $$x = y + z.$$ Equation 3 (from the first and third numerators): $$xy + zx - x^2 = zx + yz - z^2 \implies xy - x^2 = yz - z^2 \implies y(x - y) = z(y - z).$$ Rewrite as: $$y(x - y) = z(y - z).$$ 5. **Using previous results:** From above, we have: $$z = x + y,$$ $$x = y + z.$$ Substitute $$z = x + y$$ into $$x = y + z$$: $$x = y + (x + y) = x + 2y,$$ which implies: $$x - x = 2y \implies 0 = 2y \implies y = 0.$$ 6. **Substitute $$y=0$$ back:** Then $$z = x + 0 = x,$$ and $$x = 0 + z = z,$$ so $$x = z.$$ 7. **Summary of relations:** $$y = 0, \quad x = z.$$ 8. **Evaluate the expressions:** We want to prove: $$x^y y^x = y^z z^y = z^x x^z.$$ Substitute $$y=0$$ and $$x=z$$: - $$x^y y^x = x^0 \cdot 0^x = 1 \cdot 0 = 0,$$ - $$y^z z^y = 0^x \cdot x^0 = 0 \cdot 1 = 0,$$ - $$z^x x^z = x^x \cdot x^x = x^{2x}.$$ Since $$x^{2x}$$ is not zero for $$x \neq 0,$$ the equality holds only if $$x=0$$ as well. 9. **Check if $$x=0$$:** If $$x=0$$ and $$y=0$$, then $$z = x + y = 0.$$ Then all terms are zero or undefined (like $$0^0$$), but by convention $$0^0=1$$ in combinatorics. Evaluate: $$x^y y^x = 0^0 \cdot 0^0 = 1 \cdot 1 = 1,$$ $$y^z z^y = 0^0 \cdot 0^0 = 1,$$ $$z^x x^z = 0^0 \cdot 0^0 = 1.$$ Thus, the equality holds. 10. **Conclusion:** Under the condition $$L = M = N,$$ the variables satisfy $$y=0$$ and $$x=z,$$ which leads to the equality: $$x^y y^x = y^z z^y = z^x x^z.$$