‌x3−3x2+3x+7=0 ‌⇒‌‌(x+1)(x2−4x+7)=0 ‌⇒‌‌x=−1‌ or ‌‌
4±√16−28
2(1)
‌⇒‌‌x=−1‌ or ‌2±√3i ‌∴‌‌α=−1,β=2+√3i‌ and ‌ ‌γ=2−√3i New root are −1−h,(2−h)+√3i and (2−h)−√3i Equation which have these roots is (x+1+h)(x+h−2−√3i) ‌(x+h−2+√3i)=0 ⇒(x+1+h)[(x+h−2)2−(√3i)2]=0 ⇒(x+1+h)[x2+(h−2)2. +‌2x(h−2)+3]=0 In this equation terms containing x2 and x should be missing. ‌∴(1+h)+2(h−2)=0‌ and ‌ ‌2(1+h)(h−2)+3+(h−2)2=0 Now, ‌