Answer :
Let
R1→ aR1 , R2→ bR2, R3→ cR3
C1→ C1 – C3 and C2→ C2– C3
R3→ R3 – ( R1 + R2))
C1→ C1 + C3 , C2→ C2 + C3
= 2ab (a+b+c)2 {(b+c)(c+a) – ab}
= 2abc (a+b+c)3
=RHS
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