Answer :


we get,


=


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


=


R2 R R1 and R3 R–  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

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