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

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

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2072 Kartik, 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-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Prove that [(a+b)² ca bc; ca (b+c)² ab; bc ab (c+a)²] (determinant) = 2abc(a + b + c)³.

## Short answer

Multiplying rows by c, a, b and removing column factors, then setting s=a+b+c, using C1−C3, C2−C3, factoring s², and R3−R1−R2 yields a determinant that simplifies to 2abc(a+b+c)³.

## Answer

Let

```text
Δ = | (a+b)²   ca      bc   |
    |  ca    (b+c)²    ab   |
    |  bc      ab    (c+a)² |
```

#### Step 1: Remove the products
Multiply R₁ by c, R₂ by a, R₃ by b. Then

```text
abc·Δ = | c(a+b)²   c²a      bc²   |
        |  ca²     a(b+c)²   a²b   |
        |  b²c      ab²    b(c+a)² |
```

Take c common from C₁, a common from C₂, b common from C₃. This gives a factor abc, which cancels with the abc on the left:

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

To keep the algebra simple put x = c, y = a, z = b, so that a + b = y + z, b + c = z + x and c + a = x + y:

```text
Δ = | (y+z)²   x²      x²    |
    |  y²    (z+x)²    y²    |
    |  z²      z²    (x+y)²  |
```

Let s = x + y + z (= a + b + c).

#### Step 2: C₁ → C₁ − C₃, C₂ → C₂ − C₃
- Column 1: (y+z)² − x² = (y+z−x)s, 0, z² − (x+y)² = (z−x−y)s
- Column 2: 0, (z+x)² − y² = (z+x−y)s, z² − (x+y)² = (z−x−y)s

Take s from each of C₁, C₂ and write p = y+z−x, q = z+x−y, r = z−x−y:

```text
Δ = s² | p  0  x² |
       | 0  q  y² |
       | r  r  (x+y)² |
```

#### Step 3: R₃ → R₃ − R₁ − R₂
- r − p = −2y, r − q = −2x, (x+y)² − x² − y² = 2xy

```text
Δ = s² | p    0    x²  |
       | 0    q    y²  |
       | −2y −2x   2xy |
```

#### Step 4: Expand along R₁
```text
Δ/s² = p(2xyq + 2xy²) + x²(2yq)
     = 2xy [ p(q + y) + xq ]
```
With q + y = z + x: p(z + x) + xq = (y+z−x)(z+x) + x(z+x−y) = yz + xy + z² − x² + xz + x² − xy = z(x + y + z) = zs.

So Δ = s² · 2xy · zs = 2xyz s³, 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. Why multiply R₁, R₂, R₃ by c, a, b?
   A. To make the determinant zero
   B. To remove the products and get squares, with abc cancelling later
   C. To transpose
   D. To swap rows
   Answer: B. Then c, a, b are factored from the columns, cancelling abc. (Key point: Removing products)

2. Substituting x=c, y=a, z=b, a+b becomes:
   A. y+z
   B. x+y
   C. x+z
   D. x
   Answer: A. a+b = y+z. (Key point: Substitution)

3. After C₁−C₃ and C₂−C₃, what factor can be taken from C₁ and C₂ each?
   A. 2s
   B. xyz
   C. 1
   D. s = x+y+z
   Answer: D. Entries like (y+z)²−x² = (y+z−x)(x+y+z). (Key point: Factor s)

4. Final value:
   A. abc(a+b+c)³
   B. 2(a+b+c)³
   C. 2abc(a+b+c)³
   D. 2abc(a+b+c)²
   Answer: C. Δ = s²·2xy·zs = 2xyz s³. (Key point: Result)

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