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

# Determinant identity: (a + b + c)³

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2080 Baishakh, 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-a-b-c-3
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Use properties of determinants to show that [a−b−c 2a 2a; 2b b−c−a 2b; 2c 2c c−a−b] (determinant) = (a + b + c)³.

## Short answer

Apply R1 → R1+R2+R3, factor (a+b+c), then C2−C1 and C3−C1 to obtain a triangular matrix. The determinant simplifies to (a+b+c)³.

## Answer

Let s = a + b + c and

```text
Δ = | a−b−c    2a      2a   |
    |  2b    b−c−a     2b   |
    |  2c      2c    c−a−b  |
```

**Step 1: R₁ → R₁ + R₂ + R₃.** The new first row is:

- column 1: (a − b − c) + 2b + 2c = a + b + c
- column 2: 2a + (b − c − a) + 2c = a + b + c
- column 3: 2a + 2b + (c − a − b) = a + b + c

Take s common from R₁:

```text
Δ = s | 1      1       1   |
      | 2b   b−c−a     2b  |
      | 2c     2c    c−a−b |
```

**Step 2: C₂ → C₂ − C₁, C₃ → C₃ − C₁.**

- R₂: b − c − a − 2b = −(a + b + c) = −s, and 2b − 2b = 0
- R₃: 2c − 2c = 0, and c − a − b − 2c = −s

```text
Δ = s | 1   0    0  |
      | 2b −s    0  |
      | 2c  0   −s  |
```

The matrix is lower triangular: determinant = 1 · (−s) · (−s) = s².

Δ = s · s² = **(a + b + c)³.** ∎

## Check yourself

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

1. Which operation gives a common factor?
   A. R1 -> R1 + R2 + R3
   B. R1 -> R1 - R2
   C. C1 -> C1 + C2
   D. R2 -> R2 + R3
   Answer: A. Row 1 becomes (s, s, s). (Key point: Row operation)

2. After C2->C2-C1, C3->C3-C1 the matrix is
   A. upper triangular with s, s, s
   B. lower triangular with diagonal 1, -s, -s
   C. identity
   D. zero
   Answer: B. The diagonal product is s^2. (Key point: Triangular form)

3. The determinant equals
   A. (a+b+c)^2
   B. -(a+b+c)^3
   C. (a+b+c)^3
   D. abc
   Answer: C. s times s^2. (Key point: Result)

## Close variants set in other exams

- [Determinant identity proof](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-proof-seven) — Mathematics III, chapter 1, asked 6×
- [Determinant identity: 2abc(a + b + c)³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-2abc-a-b-c-two) — Mathematics III, chapter 1, asked 3×
- [Determinant identity proof](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-proof) — Mathematics III, chapter 1, asked 6×
- [Determinant identity: 2(a+b+c)³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-2-a-b-c) — Mathematics III, chapter 1, asked 6×

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