| ‌=1(−3−3λ−2)+(1−λ)(4+(1+λ)(3−λ)) ‌=(−3λ−5)+(1−λ)(4+3−λ+3λ−λ2) ‌=(−3λ−5)+(1−λ)(7+2λ−λ2) ‌=(−3λ−5)+(7+2λ−λ2−7λ−2λ2+λ3) ‌=λ3−3λ2−8λ+2 ‌∴λ3−3λ2−8λ+2=Aλ3+Bλ2+Cλ+D On comparing both sides, we get A‌=1,B=−3,C=−8,D=2 ∴‌‌D+A‌=2+1=3