Question

Prove that:

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

Answer 1

`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)]`

L.H.S = `[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)]`

Apply C₁ → C₁ + C₂ + C₃

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

Taking (a + b + c) common from C₁ we get,

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

Applying, R₃ → R₃ − 2R₁

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

= (a + b + c)[(b − c)(a + b − 2c) − (c − a)(c + a − 2b)]

= a³ + b³ + c³ − 3abc

As, L.H.S = R.H.S

Hence, proved.