Prove the following identities:
(CBSE 2014)
,
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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(CBSE 2013,2014,2015)
(CBSE 2014)
(CBSE 2014)