Q. 64.7( 3 Votes )

Using the propert

Answer :


By applying R1 R1 + R2 + R3, we get




Taking (a + b +c) common from the first row, we get



By applying C2 C2 – C1, we get




By applying C3 C3 – C1, we get




Now, expanding along first row, we get


= (a + b+ c)[1×{-(a + b + c)×{-(a + b + c)} – 0}]


= (a + b + c)[(a + b + c)2]


= (a + b + c)3


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