If u = log ( x³ + y³ + z³ - 3xyz) (∂/∂x + ∂/∂y + ∂/∂z)² u = 9 / (x +y +z)² (bsc 1st semester)

1. If u = log ( x³ + y³ + z³ - 3xyz)

(∂/∂x + ∂/∂y + ∂/∂z)² u = 9 / (x +y +z)²

(∂ = Dell)

Sol: 

u = log ( x³ + y³ + z³ - 3xyz)

∂u/∂x = 3x² - 3yz / x³ + y³ + z³ - 3xyz

∂u/∂y = 3y² - 3zx / x³ + y³ + z³ - 3xyz

∂u/∂z = 3z² - 3xy / x³ + y³ + z³ - 3xyz

∂u/∂x + ∂u/∂y + ∂u/∂z = 

3x² - 3yz+ 3y² - 3zx+ 3z² - 3xy / x³ + y³ + z³ - 3xyz

∂u/∂x + ∂u/∂y + ∂u/∂z = 

3x² +3y²+3z² - 3yz - 3zx - 3xy/ x³ + y³ + z³ - 3xyz

∂u/∂x + ∂u/∂y + ∂u/∂z = 

3(x² +y²+z² - yz - zx - xy) / (x+y+z)(x² +y²+z² - yz - zx - xy)

∂u/∂x + ∂u/∂y + ∂u/∂z = 3 / (x+y+z)

Now ,

(∂/∂x + ∂/∂y + ∂/∂z)² u =

(∂u/∂x + ∂u/∂y + ∂u/∂z) × (∂u/∂x + ∂u/∂y + ∂u/∂z) 

(∂/∂x + ∂/∂y + ∂/∂z)² u =

3/ x+y+z × 3/ x+y+z = 9 / (x +y +z)²

Proved that 

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