(∂/∂x+∂/∂y+∂/∂z)² of log(x³+y³+z³−3xyz)

Short answer

Factor x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²−xy−yz−zx). Let D = ∂/∂x+∂/∂y+∂/∂z. Compute D u = 3/(x+y+z), then D² u = −9/(x+y+z)².

If u = log(x³+y³+z³−3xyz), show that (∂/∂x + ∂/∂y + ∂/∂z)² u = −9/(x+y+z)².

Let u = log(x³ + y³ + z³ − 3xyz) and D = ∂/∂x + ∂/∂y + ∂/∂z.

x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx).

Write v = x³ + y³ + z³ − 3xyz, s = x + y + z, q = x² + y² + z² − xy − yz − zx, so v = s q.

∂v/∂x = 3x² − 3yz,  ∂v/∂y = 3y² − 3zx,  ∂v/∂z = 3z² − 3xy
Dv = 3(x² + y² + z² − xy − yz − zx) = 3q

Therefore

Du = Dv / v = 3q/(s q) = 3/(x + y + z).

Since D(x + y + z) = 1 + 1 + 1 = 3,

D²u = D( 3 (x + y + z)⁻¹ )
    = −3 (x + y + z)⁻² · D(x + y + z)
    = −3 · 3 / (x + y + z)²
    = −9/(x + y + z)²

Hence (∂/∂x + ∂/∂y + ∂/∂z)² u = −9/(x + y + z)². ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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