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Question-101514

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Question Number 101514 by bemath last updated on 03/Jul/20
Commented by Dwaipayan 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
(a+b+c)2=9a2+b2+c2+2ab+2bc+2ac=9a2+b2+c2−ab−bc−ca=3(a+b+c)(a2+b2+c2−ab−bc−ca)=3.3=9a3+b3+c3−3abc=9
Commented by Rasheed.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).
(a+b+c)2=9a2+b2+c2+2ab+2bc+2ac=9…(I)a2+b2+c2−ab−bc−ca=3……(II)(a+b+c)(a2+b2+c2−ab−bc−ca)=3.3=9a3+b3+c3−3abc=9Plinsertsomestepsbetween(I)&(II).
Commented by mr 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)
a2+b2+c2+2ab+2bc+2ac=9…(I)a2+b2+c2+2ab+2bc+2ac−6=9−6a2+b2+c2+2ab+2bc+2ac−3(ab+bc+ca)=9−6a2+b2+c2−ab−bc−ca=3……(II)
Commented by Rasheed.Sindhi last updated on 03/Jul/20
Thamks sir mr W!
ThamkssirmrW!
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 ■
a+b+c=3,ab+bc+ca=2a3+b3+c3−3abc=?Inordertodetermineaboveweneeda2+b2+c2(a+b+c)2=(3)2a2+b2+c2+2(ab+bc+ca)=9a2+b2+c2+2(2)=9a2+b2+c2=5Now,a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−(ab+bc+ca))=(3)(5−2)=9◼
Commented by bemath last updated on 03/Jul/20
great sir. thank you
greatsir.thankyou
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
a3+b3+c3=(a+b)3−3ab(a+b)+c3−3abc=[(3−c)3+c3]−3ab(a+b+c)=(3+c−c)[(3−c)2−(3−c)c+c2]−9ab=3(9−6c+c2−3c+c2+c2)−9ab=3(9−9c+3c2)−9ab=9(c2−3c+3−ab)=9[c(−a−b)+3−ab]=9[3−(ab+bc+ca)]=9(−2+3)=9otherway:a3+b3+c3−3abc=(a+b+c)3−3(a+b)(a+c)(b+c)−3abc=33−3[(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 by Rasheed.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
Firstline:a3+b3+c3−3abc=(a+b)3−3ab(a+b)+c3−3abc
Commented by 1549442205 last updated on 05/Jul/20
Thank you sir.Excuse me as missed ”−3abc”
Thankyousir.Excusemeasmissed″−3abc″

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