Q. 183.6( 5 Votes )

Prove the followi

Answer :


Taking, a,b,c common from C1, C2, C3 respectively we get,



Applying, C1C1 + C2 + C3, we get,




Applying, C2C2 – C1 and C3C3 – C1, we get,



Applying, C1C1 + C2 + C3, we get,



Taking c, a, b common from C1, C2, C3 respectively, we get,



Applying, R3R3 – R1, we have



= 2a2b2c2(2)


= 4a2b2c2 = R.H.S


Hence, proved.

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