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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
$$\left({a}+{b}+{c}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ac}=\mathrm{9} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}=\mathrm{3} \\ $$$$\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)=\mathrm{3}.\mathrm{3}=\mathrm{9} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\mathrm{9} \\ $$
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).
$$\left({a}+{b}+{c}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ac}=\mathrm{9}…\left(\mathrm{I}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}=\mathrm{3}……\left(\mathrm{II}\right) \\ $$$$\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)=\mathrm{3}.\mathrm{3}=\mathrm{9} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\mathrm{9} \\ $$$${Pl}\:{insert}\:{some}\:{steps}\:{between}\:\left(\mathrm{I}\right) \\ $$$$\&\:\left(\mathrm{II}\right). \\ $$
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)
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ac}=\mathrm{9}…\left(\mathrm{I}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ac}−\mathrm{6}=\mathrm{9}−\mathrm{6} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ac}−\mathrm{3}\left({ab}+{bc}+{ca}\right)=\mathrm{9}−\mathrm{6} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}=\mathrm{3}……\left(\mathrm{II}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 03/Jul/20
Thamks sir mr W!
$$\mathcal{T}{hamks}\:{sir}\:{mr}\:{W}! \\ $$
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 ■
$${a}+{b}+{c}=\mathrm{3},{ab}+{bc}+{ca}=\mathrm{2} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=? \\ $$$${In}\:{order}\:{to}\:{determine}\:{above}\:{we} \\ $$$${need}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} =\left(\mathrm{3}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right)=\mathrm{9} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}\right)=\mathrm{9} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{5} \\ $$$$\mathcal{N}{ow}, \\ $$$$\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc} \\ $$$$\:\:=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\left({ab}+{bc}+{ca}\right)\right) \\ $$$$\:=\left(\mathrm{3}\right)\left(\mathrm{5}−\mathrm{2}\right)=\mathrm{9}\:\blacksquare \\ $$
Commented by bemath last updated on 03/Jul/20
great sir. thank you
$$\mathrm{great}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$
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
$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} =\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{3}} −\mathrm{3ab}\left(\mathrm{a}+\mathrm{b}\right)+\mathrm{c}^{\mathrm{3}} −\mathrm{3abc} \\ $$$$=\left[\left(\mathrm{3}−\mathrm{c}\right)^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \right]−\mathrm{3ab}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)= \\ $$$$\left(\mathrm{3}+\mathrm{c}−\mathrm{c}\right)\left[\left(\mathrm{3}−\mathrm{c}\right)^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{c}\right)\mathrm{c}+\mathrm{c}^{\mathrm{2}} \right]−\mathrm{9ab} \\ $$$$=\mathrm{3}\left(\mathrm{9}−\mathrm{6c}+\mathrm{c}^{\mathrm{2}} −\mathrm{3c}+\mathrm{c}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)−\mathrm{9ab} \\ $$$$=\mathrm{3}\left(\mathrm{9}−\mathrm{9c}+\mathrm{3c}^{\mathrm{2}} \right)−\mathrm{9ab}=\mathrm{9}\left(\mathrm{c}^{\mathrm{2}} −\mathrm{3c}+\mathrm{3}−\mathrm{ab}\right) \\ $$$$=\mathrm{9}\left[\mathrm{c}\left(−\mathrm{a}−\mathrm{b}\right)+\mathrm{3}−\mathrm{ab}\right]=\mathrm{9}\left[\mathrm{3}−\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)\right] \\ $$$$=\mathrm{9}\left(−\mathrm{2}+\mathrm{3}\right)=\mathrm{9} \\ $$$$\boldsymbol{\mathrm{other}}\:\boldsymbol{\mathrm{way}}:\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\boldsymbol{\mathrm{c}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{\mathrm{abc}}=\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{3}} \\ $$$$−\mathrm{3}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)−\mathrm{3}\boldsymbol{\mathrm{abc}}= \\ $$$$\mathrm{3}^{\mathrm{3}} −\mathrm{3}\left[\left(\mathrm{3}−\boldsymbol{\mathrm{a}}\right)\left(\mathrm{3}−\boldsymbol{\mathrm{b}}\right)\left(\mathrm{3}−\boldsymbol{\mathrm{c}}\right)\right]−\mathrm{3}\boldsymbol{\mathrm{abc}}= \\ $$$$\mathrm{27}−\mathrm{3}\left[\mathrm{27}−\mathrm{9}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)+\mathrm{3}\left(\boldsymbol{\mathrm{ab}}+\boldsymbol{\mathrm{bc}}+\boldsymbol{\mathrm{ca}}\right)−\boldsymbol{\mathrm{abc}}\right]−\mathrm{3}\boldsymbol{\mathrm{abc}} \\ $$$$=\mathrm{27}−\mathrm{3}\left(\mathrm{27}−\mathrm{9}.\mathrm{3}+\mathrm{3}.\mathrm{2}−\boldsymbol{\mathrm{abc}}\right)−\mathrm{3}\boldsymbol{\mathrm{abc}} \\ $$$$=\mathrm{27}−\mathrm{3}\left(\mathrm{6}−\boldsymbol{\mathrm{abc}}\right)−\mathrm{3}\boldsymbol{\mathrm{abc}}=\mathrm{27}−\mathrm{18}=\mathrm{9} \\ $$
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
$${First}\:{line}: \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} −\mathrm{3}{abc}=\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{3}} −\mathrm{3ab}\left(\mathrm{a}+\mathrm{b}\right)+\mathrm{c}^{\mathrm{3}} −\mathrm{3abc} \\ $$$$ \\ $$
Commented by 1549442205 last updated on 05/Jul/20
Thank you sir.Excuse me as missed ”−3abc”
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\mathrm{Excuse}\:\mathrm{me}\:\mathrm{as}\:\mathrm{missed}\:''−\mathrm{3abc}'' \\ $$

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