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You study in ninth standard?
let f(x)=x2013
x2-1 can be resolved into factors as (x+1)(x-1)
now x+1=0 => x=-1
substitue x=-1 in f(x)
f(-1)= (-1)2013 =-1
then x-1=0 => x=1
 sub x=1 in f(x)
f(1) = (1)2013 =1
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