Answer :

Given,

LHS =


RHS = 4a2b2c2


LHS = Δ =


Taking out common factors a, b and c from C1, C2 and C3, we have


Δ = abc


Applying Elementary Transformations


R2 R2 – R1 and R3 R3 – R1


Δ = abc


R2 R2 + R1


Δ = abc


R3 R3 + R2


Δ = abc


Δ = 2ab2c


C2 C2 – C1


Δ = 2ab2c


Expanding along R3 we get,


Δ = 2ab2c [a (c – a) + a (a + c)]


= 2ab2c [ac – a2 + a2 + ac]


= 2ab2c (2ac)


= 4a2b2c2


Δ = RHS


LHS = RHS


Hence, Proved


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