Q. 194.0( 5 Votes )

Using properties

Answer :


Given,



Performing the operation C1 C1 – C2 and C2 C2 – C3, we get



We know that a3 – b3 = (a – b) (a2 + ab + b2)



Taking (a – b) and (b – c) common from C1 and C2 respectively, we get



Expanding along R1, we get


= (a – b) (b – c) [(b2 + bc + c2) – (a2 + ab + b2)]


= (a – b) (b – c) [(c2 – a2) + (bc – ab)]


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


= (a – b) (b – c) (c – a) (c + a + b)


= (a – b) (b – c) (c – a) (a + b + c)


= RHS


Hence proved.


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