Given: (a + b + c) = 7 a2+b2+c2=17 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: We have (a + b + c) = 7 Squaring both side, we get (a+b+c)2=72 ⇒ a2+b2+c2+2ab+2bc+2ca=49 ⇒ 17+2(ab+bc+ca)=49 ⇒ 2(ab+bc+ca)=32 ⇒ (ab+bc+ca)=16 Now, we have to find the value of a3+b3+c3−3abc a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca) ⇒ a3+b3+c3−3abc=7×(17−16) ⇒ a3+b3+c3−3abc=7 ∴ The value of a3+b3+c3−3abc is 7.