Given: a+b+c=4,abc=−30 and a3+b3+c3=106 Formula Used: a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca) (a+b+c)2=(a2+b2+c2+2ab+2bc+2ca) Calculation: According to the question 106−3×(−30)=4(a2+b2+c2−ab−bc−ca) ⇒ (a2+b2+c2−ab−bc−ca)=49 ...(i) (a+b+c)2=(a2+b2+c2+2ab+2bc+2ca) ⇒ 42=(a2+b2+c2+2ab+2bc+2ca) ⇒ (a2+b2+c2+2ab+2bc+2ca)=16 ...(ii) From (1) and (2), we get 3(ab + bc + ca) = 16 – 49 ∴ The value of (ab + bc + ca) is -11.