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

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

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2082 Shrawan, 5 marks; 2081 Baishakh, 5 marks; 2070 Chaitra, 5 marks (asked 3 times)
- 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

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

## Short answer

Factor s = a+b+c after column operations, then apply row operations to get Δ = s² times a simpler determinant. Expanding and simplifying gives Δ = 2abc(a+b+c)³.

## Answer

Let Δ be the given determinant and s = a + b + c.

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

#### Step 1: C₁ → C₁ − C₃, C₂ → C₂ − C₃
Using x² − y² = (x − y)(x + y):

- Column 1: (b+c)² − a² = (b+c−a)s, 0, c² − (a+b)² = (c−a−b)s
- Column 2: 0, (c+a)² − b² = (c+a−b)s, c² − (a+b)² = (c−a−b)s

Take s out of each of C₁ and C₂ (factor s²). Write p = b+c−a, q = c+a−b, r = c−a−b:

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

#### Step 2: R₃ → R₃ − R₁ − R₂
- First entry: r − p = (c−a−b) − (b+c−a) = −2b
- Second entry: r − q = (c−a−b) − (c+a−b) = −2a
- Third entry: (a+b)² − a² − b² = 2ab

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

#### Step 3: Expand along R₁
```text
Δ/s² = p(2abq + 2ab²) + a²(2bq)
     = 2ab[ p(q + b) + a q ]
```
Now q + b = c + a, so
p(q + b) + aq = (b+c−a)(c+a) + a(c+a−b)
= (bc + ab + c² − a²) + (ac + a² − ab) = c(a + b + c) = cs.

Therefore Δ/s² = 2ab · cs, i.e.

**Δ = 2abc(a + b + c)³.** ∎

## Check yourself

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

1. Which operations start the reduction?
   A. R1 -> R1 + R2 + R3
   B. C1 -> C1 + C2 + C3
   C. Expand along row 1 immediately
   D. C1 -> C1 - C3 and C2 -> C2 - C3
   Answer: D. The differences of squares then give common factor s = a+b+c in both columns. (Key point: First column operations)

2. After the column operations, what factor is taken out?
   A. s^2 where s = a + b + c
   B. s
   C. abc
   D. 2s
   Answer: A. Both C1 and C2 contain the factor s, giving s^2. (Key point: Common factor)

3. Which operation creates simple entries in the third row?
   A. R3 -> R3 + R1 + R2
   B. R3 -> R3 - R1 - R2
   C. R1 -> R1 - R3
   D. R3 -> 2R3
   Answer: B. It gives -2b, -2a, 2ab, which makes the expansion easy. (Key point: Row operation)

4. What is the final value of the determinant?
   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. s^2 times 2ab*cs gives 2abc s^3. (Key point: Final result)

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