1. **Problem statement:** Given the function $$U = \log(x^3 + y^3 + z^3 - 3xyz),$$ prove that $$\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)^2 U = -\frac{9}{(x + y + z)^2}.$$ 2. **Recall the formula and rules:** - The operator $$\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)^2$$ means applying the sum of partial derivatives twice. - Use chain rule and implicit differentiation. 3. **Step 1: Define $$S = x^3 + y^3 + z^3 - 3xyz$$ so that $$U = \log S.$$** 4. **Step 2: Compute first derivatives:** $$\frac{\partial U}{\partial x} = \frac{1}{S} \frac{\partial S}{\partial x}, \quad \frac{\partial U}{\partial y} = \frac{1}{S} \frac{\partial S}{\partial y}, \quad \frac{\partial U}{\partial z} = \frac{1}{S} \frac{\partial S}{\partial z}.$$ Calculate $$\frac{\partial S}{\partial x}$$: $$\frac{\partial S}{\partial x} = 3x^2 - 3yz.$$ Similarly, $$\frac{\partial S}{\partial y} = 3y^2 - 3xz,$$ $$\frac{\partial S}{\partial z} = 3z^2 - 3xy.$$ 5. **Step 3: Sum of first derivatives:** $$\frac{\partial U}{\partial x} + \frac{\partial U}{\partial y} + \frac{\partial U}{\partial z} = \frac{1}{S} \left(3x^2 - 3yz + 3y^2 - 3xz + 3z^2 - 3xy\right).$$ Group terms: $$= \frac{3}{S} \left(x^2 + y^2 + z^2 - xy - yz - zx\right).$$ 6. **Step 4: Note the identity:** $$x^2 + y^2 + z^2 - xy - yz - zx = \frac{1}{2} \left((x - y)^2 + (y - z)^2 + (z - x)^2\right) \geq 0.$$ 7. **Step 5: Compute the second derivative operator:** We want $$\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)^2 U = \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right) \left(\frac{\partial U}{\partial x} + \frac{\partial U}{\partial y} + \frac{\partial U}{\partial z}\right).$$ Let $$V = \frac{\partial U}{\partial x} + \frac{\partial U}{\partial y} + \frac{\partial U}{\partial z} = \frac{3}{S} (x^2 + y^2 + z^2 - xy - yz - zx).$$ 8. **Step 6: Apply the operator to $$V$$:** $$\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right) V = \frac{\partial V}{\partial x} + \frac{\partial V}{\partial y} + \frac{\partial V}{\partial z}.$$ 9. **Step 7: Compute $$\frac{\partial V}{\partial x}$$:** Using quotient rule, $$\frac{\partial V}{\partial x} = 3 \frac{\partial}{\partial x} \left( \frac{Q}{S} \right),$$ where $$Q = x^2 + y^2 + z^2 - xy - yz - zx.$$ Calculate $$\frac{\partial Q}{\partial x}$$: $$\frac{\partial Q}{\partial x} = 2x - y - z.$$ Calculate $$\frac{\partial S}{\partial x} = 3x^2 - 3yz$$ (from step 4). By quotient rule: $$\frac{\partial}{\partial x} \left( \frac{Q}{S} \right) = \frac{S (2x - y - z) - Q (3x^2 - 3yz)}{S^2}.$$ 10. **Step 8: Similarly compute $$\frac{\partial V}{\partial y}$$ and $$\frac{\partial V}{\partial z}$$:** By symmetry, $$\frac{\partial V}{\partial y} = 3 \frac{S (2y - x - z) - Q (3y^2 - 3xz)}{S^2},$$ $$\frac{\partial V}{\partial z} = 3 \frac{S (2z - x - y) - Q (3z^2 - 3xy)}{S^2}.$$ 11. **Step 9: Sum all partial derivatives:** $$\frac{\partial V}{\partial x} + \frac{\partial V}{\partial y} + \frac{\partial V}{\partial z} = \frac{3}{S^2} \left[ S \sum (2x - y - z) - Q \sum (3x^2 - 3yz) \right].$$ Note that $$\sum (2x - y - z) = (2x - y - z) + (2y - x - z) + (2z - x - y) = 0,$$ and $$\sum (3x^2 - 3yz) = 3(x^2 + y^2 + z^2) - 3(xy + yz + zx) = 3Q.$$ Therefore, $$\frac{\partial V}{\partial x} + \frac{\partial V}{\partial y} + \frac{\partial V}{\partial z} = \frac{3}{S^2} \left[ 0 - Q (3Q) \right] = -\frac{9 Q^2}{S^2}.$$ 12. **Step 10: Recall the identity:** $$Q = x^2 + y^2 + z^2 - xy - yz - zx = \frac{1}{2} ((x - y)^2 + (y - z)^2 + (z - x)^2).$$ Also, from the factorization of $$S$$: $$S = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = (x + y + z) Q.$$ Thus, $$S = (x + y + z) Q \implies S^2 = (x + y + z)^2 Q^2.$$ 13. **Step 11: Substitute into the expression:** $$-\frac{9 Q^2}{S^2} = -\frac{9 Q^2}{(x + y + z)^2 Q^2} = -\frac{9}{(x + y + z)^2}.$$ 14. **Final result:** $$\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)^2 U = -\frac{9}{(x + y + z)^2}.$$ This completes the proof.