Question

Using properties of determinants prove that:

`|(a - b, b + c, a),(b - c, c + a, b),(c - a, a + b, c)| = a^3 + b^3 + c^3 - 3abc`.

Answer 1

To Prove

`|(a - b, b + c, a),(b - c, c + a, b),(c - a, a + b, c)| = a^3 + b^3 + c^3 - 3abc`

From L.H.S. `|(a - b, b + c, a),(b - c, c + a, b),(c - a, a + b, c)|`

R₁ → R₁ + R₂ + R₃

= `|(0, 2(a + b + c), a + b + c),(b - c, c + a, b),(c - a, a + b, c)|`

Taking (a + b + c) common from R₁

= `(a + b + c)|(0, 2, 1),(b - c, c + a, b),(c - a, a + b, c)|`

C₂ → C₂ – 2C₃

= `(a + b + c)|(0, 0, 1),(b - c, c + a - 2b, b),(c - a, a + b - 2c, c)|`

Expand along R₁

= (a + b + c)[0 – 0 + l{(b – c)(a + b – 2c) – (c – a)(c + a – 2b)}]

= (a + b + c)[(ab + b² – 2bc – ac – bc + 2c²) – (c² + ac – 2bc – ac – a² + 2ab]

= (a + b + c)[ab + b² – 3bc – ac + 2c² – c² + 2bc + a² – 2ab]

= (a + b + c)[a² + b² + c² – ab – bc – ac]

= a³ + b³ + c³ – 3abc   ...[Using identity]

= R.H.S.

Hence Proved.