Question

 (n^2-1)*(n^4+n^2+1)-(n^2-1)^3

Answer 1

n^6+n^4+n^2-n^4-n^2-1-n^6-3n^4+3n^2-1=
=-3n^4+3n^2-2

Answer 2

$\(n^2-1)*(n^4+n^2+1)-(n^2-1)^3= \. \(n^2-1)[(n^4+n^2+1)-(n^2-1)^2]= \. \(n^2-1)(n^4+n^2+1-n^4+2n^2-1))= \. \3n^2(n^2-1)=3n^4-3n^2$