Answer :

we get,

=

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

=

R_{2}→ R_{2 }– R_{1} and R_{3}→ R_{3 }– R_{1} we get,

=

Taking 2 common from R_{2} and R_{3}

=

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

= 4(a + b + c)(a^{2} + 2ac + 2ac + 4bc–a^{2} + 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

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