Question

Using the property of determinants and without expanding, prove that

|(a - b - c, 2a, 2a), (2b, b - c - a, 2b), (2c, 2c, c - a - b)| = (a + b + c)³

Answer 1

(Using R₁ → R₁ + R₂ + R₃)

Take out (a + b + c) common from R₁, we get

(Using C₂ → C₂ - C₁ and C₃ → C₃ - C₁)

Expanding along R₁, we get 

= (a + b + c) {1(− b − c − a) (− c − a − b)} 

= (a + b + c) [− (b + c + a) × ( −) (c + a + b)]

= (a + b + c)(a + b + c)(a + b + c) = (a + b + c)³ = RHS