Let (a+x)n= Odd trems(A) + Even terms (B)So (a−x)n= Odd terms(A) - Even terms(B)∴(a+x)n−(a−x)n=(A+B)−(A−B)=2B=2[ even terms] =2[T2+T4+T6+⋯] So in case of (3+1)2n−(3−1)2n=2[T2+T4+T6+⋯]=2[2nC1⋅(3)2n−1+2nC3⋅(3)2n−3+⋯]Here in (3)2n−1,2n−1 is odd number. So there will be always 3 in (3)2n−1. So (3+1)2n−(3−1)2n will be always irrational number.