f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3) ⇒‌‌f′(x)=3x2+2xf′(1)+f′′(2)...(i) ⇒‌‌f′′(x)=6x+2f′(1)..(ii) ⇒‌‌f′′′(x)=6...(iii) Put x=1, in Eq. (i) f′(1)=3+2f′(1)+f′′(2)...(iv) Put x=2 in Eq. (ii) f′′(2)=12+2f′(1)....(v) From Eqs. (iv) and (v), ‌−3−f′(1)=12+2f′(1) ⇒3f′(1)=−15 ⇒f′(1)=−5 ⇒f′′(2)=2‌‌[‌ from Eq. (v)] ‌ Put x=3 in Eq. (iii), we get ‌f′′′(3)=6 ‌f(x)=x3−5x2+2x+6 ‌f(2)=8−20+4+6=−2