Answer :
,
R.H.S = 2(a + b + c)2
Applying C1→C1 + C2 + C3, we have
Taking, 2(a + b + c) common we get,
Now, applying R2→R2 – R1 and R3→R3 – R1, we get,
Thus, we have
L.H.S = 2(a + b + c)[1(a + b + c)2]
= 2(a + b + c)3 = R.H.S
Hence, proved.
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