Sửa đề: a^3+b^3+c^3=3abc

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

=>ĐPCM

`a^3+b^3+c^3=3abc(***)`

`a^3+b^3+c^3-3abc=0`

`<=>a^3+3ab(a+b)+c^3-3ab(a+b)-3abc=0`

`<=>(a+b)^3+c^3-3ab(a+b+c)=0`

`<=>(a+b+c)(a^2+b^2+2ab-ac-bc)-3ab(a+b+c)=0`

`<=>(a+b+c)(a^2+b^2+c^2-ac-bc-ab)=0`

Luôn đúng với `a+b+c=0`

`=>(***)` được chứng minh.

Ta có: \(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^3=\left(-c\right)^3\)

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3+c^3=0\)

\(\Leftrightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)(đpcm)

\(GT\Rightarrow a+b=-c\)

Ta có \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(-c\right)^3+c^3-3ab\left(-c\right)-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\) Vậy...

Ta có \(a^3+b^3+c^3-3abc=a^3+b^3+c^3-abc-abc-abc+ac^2+a^2c-ac^2-a^2c+ab^2+a^2b-ab^2-a^2b+b^2c+bc^2-b^2c-bc^2=a^2\left(a+b+c\right)+b^2\left(a+b+c\right)+c^2\left(a+b+c\right)-ab\left(a+b+c\right)-bc\left(a+b+c\right)-ac\left(a+b+c\right)=\frac{1}{2}\left(a+b+c\right)\left(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right)\)=0 vạy a+b+c=0

ta có a+b+c=0=>a+b=-c

ta lại có a^3+b^3+c^3

          =(A+b)(a^2-ab+b^2)+c^3

          =-c [(A+b)^2-2ab-ab)]+c^3

        =   -c (-c^2-3ab)+c^3

        =      -c(c^2-3ab)+c^3

         =  -c^3 +3abc+c^3

         =3abc

vì mọi số mũ abc đều mũ 3 nên 3abc là kết quả khi cộng các số đó mũ 3 thì kết quả ko thay đổi

a+b+c=0

=>(a+b+c)³=0

=>a³+b³+c³+3a²b+3ab²+3b²c+3bc²+3a²c+3ac²+6abc=0

=>a³+b³+c³+(3a²b+3ab²+3abc)+(3b²c+3bc²+3abc)+(3a²c+3ac²+3abc)-3abc=0

=>a³+b³+c³+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc

Do a+b+c=0

=>a³+b³+c³=3abc(ĐPCM)

 Ta có :(a+b+c)³=a³+b³+c³+3a²b+3a²c+3b²c+3b²a+3c²a+3c²b+6abc

            (a+b+c)³=a³+b³+c³+(3a²b+3a²b+3abc)+(3b²c+3b²a+3abc)+(3c²a+3c²b+3abc)-3abc

            (a+b+c)³=a³+b³+c³+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)-3abc

            (a+b+c)³=a³+b³+c³+3(a+b+c)(ab+bc+ac)-3abc

  thay a+b+c=0 ta được 

              0³=a³+b³+c³+3.0(ab+bc+ac)-3abc

             0=a³+b³+c³-3abc

=>a³+b³+c³=3abc

Có nhiều cách để chứng minh. Chẳng hạn, thay a^3 +b^3 =(a+b)^3 -3ab(a+b) và a + b = -c, ta được

a^3 + b^3 + c^3 = (a+b)^3 - 3ab(a+b) + c^3 = -c^3 - 3ab(-c) + c^3 =3abc

Ta có :(a+b+c)³=a³+b³+c³+3a²b+3a²c+3b²c+3b²a+3c²a+3c²b+6abc

            (a+b+c)³=a³+b³+c³+(3a²b+3a²b+3abc)+(3b²c+3b²a+3abc)+(3c²a+3c²b+3abc)-3abc

            (a+b+c)³=a³+b³+c³+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)-3abc

            (a+b+c)³=a³+b³+c³+3(a+b+c)(ab+bc+ac)-3abc

  thay a+b+c=0 ta được 

              0³=a³+b³+c³+3.0(ab+bc+ac)-3abc

             0=a³+b³+c³-3abc

=>a³+b³+c³=3abc

a+b+c=0

=>a+b=-c

=>(a+b)³=-c³

=>a³+b³+3a²b+3ab²+c³=0

=>a³+b³+c³+3ab(a+b)=0

Mà a+b=-c

=>a³+b³+c³+3ab.(-c)=0

=>a³+b³+c³-3abc=0

=>a³+b³+c³=3abc

Có: \(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^3=-c^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Leftrightarrow a^3+b^3-3abc=-c^3\) (Vì a+b=-c)

\(\Leftrightarrow a^3+b^3+c^2=3abc\)

Ta có :(a+b+c)³=a³+b³+c³+3a²b+3a²c+3b²c+3b²a+3c²a+3c²b+6abc

            (a+b+c)³=a³+b³+c³+(3a²b+3a²b+3abc)+(3b²c+3b²a+3abc)+(3c²a+3c²b+3abc)-3abc

            (a+b+c)³=a³+b³+c³+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)-3abc

            (a+b+c)³=a³+b³+c³+3(a+b+c)(ab+bc+ac)-3abc

  Thay a+b+c=0 ta được 

              0³=a³+b³+c³+3.0(ab+bc+ac)-3abc

             0=a³+b³+c³-3abc

=>a³+b³+c³=3abc

1. \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(abc\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc\right)-3ab\left(a+b+c\right)\)

\(\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

2. \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

3.Còn có a + b + c = 0 nữa mà bn.

\(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)

+ \(a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\ \left(c-a\right)^2=0\end{matrix}\right.\)

\(\Rightarrow a=b=c\)

\(\frac{a^3+b^3+c^3-3abc}{a+b+c}=\frac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{a+b+c}=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a+b+c}\)

\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a+b+c}=a^2+b^2+c^2-ab-bc-ca\)

\(=\frac{1}{2}\left(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\right)\)

\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\) (đpcm)

a+b+c=0

=>(a+b+c)³=0

=>a³+b³+c³+3a²b+3ab²+3b²c+3bc²+3a²c+3ac²+6abc=0

=>a³+b³+c³+(3a²b+3ab²+3abc)+(3b²c+3bc²+3abc)+(3a²c+3ac²+3abc)-3abc=0

=>a³+b³+c³+3ab(a+b+c)+....mk phải ăn cơm rồi