Question

упростите (n^2-1)*(n^4+n^2+1)-(n^2-1)^3

Answer 1

$(n^2-1)(n^4+n^2+1)-(n^2-1)^3=\\=(n^2)^3-1^3-(n^6-3n^4+3n^2-1)=\\=n^6-1-n^6+3n^4-3n^2+1=3n^4-3n^2=3n^2cdot (n^2-1)=\\=3n^2(n-1)(n+1)\\ili$

$(n^2-1)(n^4+n^2+1)-(n^2-1)^3=\\=(n^2-1)cdot (n^4+n^2+1-(n^2-1)^2)=\\=(n^2-1)(n^4+n^2+1-n^4+2n^2-1)=\\=(n^2-1)(3n^2)=3n^2(n-1)(n+1)$