---
title: "(∂/∂x+∂/∂y+∂/∂z)² of log(x³+y³+z³−3xyz)"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:x-y-z-2-log"
source: https://www.bctengineer.com/q/semester_2/mathematics/x-y-z-2-log
site: BCT Engineering
---

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

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2072 Ashwin, 5 marks
- Video: [If u=log(x^3 + y^3 + z^3 - 3xyz)then prove that (del/del x  + de/del y+ del/del z)^2.u=-9/(x+y+z)^2](https://www.youtube.com/watch?v=RFmITs_6jXI) — LEARN MATHEMATICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/x-y-z-2-log
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

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

## 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.

## Answer

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

#### Step 1: factorise
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.

#### Step 2: compute Du

```text
∂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).

#### Step 3: apply D again
Since D(x + y + z) = 1 + 1 + 1 = 3,

```text
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. ∎

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. x³+y³+z³−3xyz factorises as:
   A. (x−y−z)(x²+y²+z²)
   B. (x+y+z)³
   C. (x+y+z)(x²+y²+z²−xy−yz−zx)
   D. (x+y+z)(x²+y²+z²+xy+yz+zx)
   Answer: C. This is the standard identity. (Key point: Factorisation)

2. What is Du = (∂/∂x+∂/∂y+∂/∂z)u?
   A. 3q/(x+y+z)
   B. 9/(x+y+z)
   C. 3/(x+y+z)
   D. 1/(x+y+z)
   Answer: C. Dv = 3q and v = s q, so Du = 3q/(s q) = 3/s. (Key point: Du)

3. What is D(x+y+z)?
   A. x+y+z
   B. 0
   C. 3
   D. 1
   Answer: C. D(x+y+z) = 1+1+1 = 3, which gives D²u = -3·3/s² = -9/s². (Key point: Second application)

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