Using properties of determinants, prove that

.

we get,

=

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

=

R₂→ R₂ – R₁ and R₃→ R₃ –  R₁ we get,

=

Taking 2 common from R₂ and R₃

=

= 4(a + b + c)((a + 2b)(a + 2c)–(a–b)(a–c))

= 4(a + b + c)(a² + 2ac + 2ac + 4bc–a² + ac + ab–bc)

= 4(a + b + c)3(ab + bc + ca) = 12(a + b + c)(ab + bc + ca)

L.H.S = R.H.S

Hence, proved