‌x(x3−6x2+11x−6)=0 ‌⇒‌‌x(x−1)(x−2)(x−3)=0 ‌⇒‌‌x=0,1,2,3 The least roots is 0 . Let the roots of x3+αx2+βx+6=0 be r1,r2,r3 The roots of x3+αx2+βx+6=0 increased by 1 are r1+1,r2+1,r3+1 One of these roots is 0 , so ri+1=0. for some i0. ⇒ri=−1 Since ri=−1 is root of ‌x3+αx2+βx+6=0 ‌⇒(−1)3+α(−1)2+β(−1)+6=0 ‌⇒−1+α−β+6=0 ‌⇒α−β+5=0
