Question Number 101514 by bemath last updated on 03/Jul/20

Commented byDwaipayan Shikari last updated on 03/Jul/20

(a+b+c)^2 =9  a^2 +b^2 +c^2 +2ab+2bc+2ac=9  a^2 +b^2 +c^2 −ab−bc−ca=3  (a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)=3.3=9  a^3 +b^3 +c^3 −3abc=9

Commented byRasheed.Sindhi last updated on 03/Jul/20

(a+b+c)^2 =9  a^2 +b^2 +c^2 +2ab+2bc+2ac=9...(I)  a^2 +b^2 +c^2 −ab−bc−ca=3......(II)  (a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)=3.3=9  a^3 +b^3 +c^3 −3abc=9  Pl insert some steps between (I)  & (II).

Commented bymr W last updated on 03/Jul/20

a^2 +b^2 +c^2 +2ab+2bc+2ac=9...(I)  a^2 +b^2 +c^2 +2ab+2bc+2ac−6=9−6  a^2 +b^2 +c^2 +2ab+2bc+2ac−3(ab+bc+ca)=9−6  a^2 +b^2 +c^2 −ab−bc−ca=3......(II)

Commented byRasheed.Sindhi last updated on 03/Jul/20

Thamks sir mr W!

Answered by Rasheed.Sindhi last updated on 03/Jul/20

a+b+c=3,ab+bc+ca=2  a^3 +b^3 +c^3 −3abc=?  In order to determine above we  need a^2 +b^2 +c^2   (a+b+c)^2 =(3)^2   a^2 +b^2 +c^2 +2(ab+bc+ca)=9  a^2 +b^2 +c^2 +2(2)=9  a^2 +b^2 +c^2 =5  Now,   a^3 +b^3 +c^3 −3abc    =(a+b+c)(a^2 +b^2 +c^2 −(ab+bc+ca))   =(3)(5−2)=9 ■

Commented bybemath last updated on 03/Jul/20

great sir. thank you

Answered by 1549442205 last updated on 07/Jul/20

a^3 +b^3 +c^3 =(a+b)^3 −3ab(a+b)+c^3 −3abc  =[(3−c)^3 +c^3 ]−3ab(a+b+c)=  (3+c−c)[(3−c)^2 −(3−c)c+c^2 ]−9ab  =3(9−6c+c^2 −3c+c^2 +c^2 )−9ab  =3(9−9c+3c^2 )−9ab=9(c^2 −3c+3−ab)  =9[c(−a−b)+3−ab]=9[3−(ab+bc+ca)]  =9(−2+3)=9  other way:a^3 +b^3 +c^3 −3abc=(a+b+c)^3   −3(a+b)(a+c)(b+c)−3abc=  3^3 −3[(3−a)(3−b)(3−c)]−3abc=  27−3[27−9(a+b+c)+3(ab+bc+ca)−abc]−3abc  =27−3(27−9.3+3.2−abc)−3abc  =27−3(6−abc)−3abc=27−18=9

Commented byRasheed.Sindhi last updated on 03/Jul/20

First line:  a^3 +b^3 +c^3 −3abc=(a+b)^3 −3ab(a+b)+c^3 −3abc

Commented by1549442205 last updated on 05/Jul/20

Thank you sir.Excuse me as missed ε−3abcε