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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