---
title: "Determinant identity: 2abc(a + b + c)³"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:determinant-identity-2abc-a-b-c-four"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-2abc-a-b-c-four
site: BCT Engineering
---

# Determinant identity: 2abc(a + b + c)³

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2071 Shrawan, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-2abc-a-b-c-four
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Show that [(b+c)² b² c²; a² (c+a)² c²; a² b² (a+b)²] (determinant) = 2abc(a + b + c)³.

## Short answer

With s=a+b+c, apply R1−R3 and R2−R3, factor s², then C3−C1−C2 and expand to get 2abc(a+b+c)³.

## Answer

Let s = a + b + c and

```text
Δ = | (b+c)²   b²      c²    |
    |  a²    (c+a)²    c²    |
    |  a²      b²    (a+b)²  |
```

**Step 1: R₁ → R₁ − R₃, R₂ → R₂ − R₃.** Using x² − y² = (x − y)(x + y):

- R₁ − R₃ = ((b+c)² − a², 0, c² − (a+b)²) = ((b+c−a)s, 0, (c−a−b)s)
- R₂ − R₃ = (0, (c+a)² − b², c² − (a+b)²) = (0, (c+a−b)s, (c−a−b)s)

Take s common from R₁ and from R₂ (factor s²). Write p = b+c−a, q = c+a−b, r = c−a−b:

```text
Δ = s² | p   0    r    |
       | 0   q    r    |
       | a²  b²  (a+b)² |
```

**Step 2: C₃ → C₃ − C₁ − C₂.**

- Row 1: r − p = (c−a−b) − (b+c−a) = −2b
- Row 2: r − q = (c−a−b) − (c+a−b) = −2a
- Row 3: (a+b)² − a² − b² = 2ab

```text
Δ = s² | p   0   −2b |
       | 0   q   −2a |
       | a²  b²  2ab |
```

**Step 3: Expand along R₁.**

```text
Δ/s² = p(2abq + 2ab²) + (−2b)(0·b² − q·a²)
     = 2ab·p(q + b) + 2a²b·q
     = 2ab [ p(q + b) + aq ]
```
With q + b = c + a:
p(c + a) + aq = (b+c−a)(c+a) + a(c+a−b) = (bc + ab + c² − a²) + (ac + a² − ab) = c(a + b + c) = cs.

So Δ/s² = 2ab · cs, i.e.

**Δ = 2abc(a + b + c)³.** ∎

(This is the transpose of the determinant |(b+c)² a² a²; b² (c+a)² b²; c² c² (a+b)²|, which has the same value.)

## Check yourself

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

1. R₁−R₃ and R₂−R₃ introduce factor:
   A. abc
   B. s = a+b+c, using x²−y² = (x−y)(x+y)
   C. a−b
   D. 2
   Answer: B. (b+c)²−a² = (b+c−a)(a+b+c). (Key point: Row operations)

2. C₃ → C₃ − C₁ − C₂ gives bottom entry:
   A. 2ab
   B. a²+b²
   C. 0
   D. (a+b)²
   Answer: A. (a+b)² − a² − b² = 2ab. (Key point: Column operation)

3. The common factor from R₁ and R₂ is s², giving Δ = s² times:
   A. a determinant of 1s
   B. the identity
   C. zero
   D. a 3×3 determinant with entries p,q,−2b,−2a,2ab
   Answer: D. The remaining determinant is expanded along R₁. (Key point: Factor out)

4. Δ equals:
   A. abc(a+b+c)³
   B. 2abc(a+b+c)²
   C. 2abc(a+b+c)³
   D. (a+b+c)³
   Answer: C. Δ/s² = 2ab·cs. (Key point: Result)

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