Answer :

–x + (x + 1) + (x + 2) – (x + 3) = 0

We will calculate LHS in every case and copare it with LHS


Taking x = 1,


= -1 + (1 + 1) + (1 + 2)-(1 + 3)


= -1 + 2 + 3-4


= 0


Taking x = 2,


= -2 + (2 + 1) + (2 + 2)-(2 + 3)


= -2 + 3 + 4-5


= 0


Taking x = 3,


= -3 + (3 + 1) + (3 + 2)-(3 + 3)


= -3 + 4 + 5-6


= 0


Taking x = 4,


= -4 + (4 + 1) + (4 + 2)-(4 + 3)


= -4 + 5 + 6-7


= 0


Taking x = 5,


= -5 + (5 + 1) + (5 + 2)-(5 + 3)


= -5 + 6 + 7-8


= 0


Taking x = -1,


= -(-1) + (-1 + 1) + (-1 + 2)-(-1 + 3)


= 1 + 0 + 1-2


= 0


Taking x = -2,


= -(-2) + (-2 + 1) + (-2 + 2)-(-2 + 3)


= 2-1 + 0-1


= 0


Taking x = -3,


= -(-3) + (-3 + 1) + (-3 + 2)-(-3 + 3)


= 3-2-1 + 0


= 0


Taking x = -4,


= -(-4) + (-4 + 1) + (-4 + 2)-(-4 + 3)


= 4-3-2 + 1


= 0


Taking x = -5,


= -(-5) + (-5 + 1) + (-5 + 2)-(-5 + 3)


= 5-4-3 + 2


= 0


As in each case in LHS = 0 so, LHS = RHS


Hence, above equation is an identity.


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