Given, n3−11n2+32n−28>0 When n=2,n3−11n2+32n−28=0 ⇒(n−2)(n2−9n+14)>28 ⇒(n−2)(n−7)(n−2)>28 For n<2,(n−2)(n−7)(n−2) is negative. For $2 For n>7,(n−2)(n−7)(n−2) is positive. When n=8,(n−2)(n−7)(n−2)=36, which is greater than 28. Least integral value of n which satisfies the inequation is 8 .