`(a+b+c)^2=3(ab+bc+ca)`

`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`

`<=>a^2+b^2+c^2=ab+bc+ca`

`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`

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

`VT>=0`

Dấu "=" xảy ra khi `a=b=c`

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

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

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

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

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

`**a+b+c=0`

`**a^2+b^2+c^2=ab+bc+ca`

`<=>a=b=c`

1) Có: \(a+b+c=0\)

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

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

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

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

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

2)Có: \(a+b-c=0\)

\(\Leftrightarrow a+b=c\)

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

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

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

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

a +b +c=0

⇔\(\left(a+b+c\right)^3\)

 ⇔\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3a^2c+3ac^2+6abc=0\)

⇔\(a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)-3abc=0\)

⇔ \(a^3+b^3+c^3+3ab\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)=3abc\)

Vì a+b+c= 0

⇒\(a^3+b^3+c^3=3abc\)

Chúc bạn học tốt!

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

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\) (chuyển vế qua)

\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

Do VP >=0 với mọi a, b, c. Nên để đăng thức xảy ra thì a = b = c

c) a + b + c = 0 suy ra a = -(b+c)

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

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

\(=3bc.\left[-\left(b+c\right)\right]=3abc\) (đpcm)

a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

Do VT >=0 với mọi a, b, c nên a = b = c 1

tí đăng tiếp

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

\(=0+3abc=3abc\)

Lời giải:

Từ $a+b+c=0\Rightarrow a+b=-c$

Theo HĐT đáng nhớ:

\((a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab(a+b)\)

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

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

Ta có đpcm.

ta có a+b+c=0

  =>a+b=-c

ta có a^3 +b^3+c^3

      =(a+b)(a^2-ab+b^2)+c^3

      =-c(a^2+b^2-ab)+c^3

      =-c[(a+b)^2-2ab-ab]+c^3

       = -c[(-c)^2-3ab]+c^3

       = (-c)^3+3abc+c^3

         =3abc

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

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

\(\Leftrightarrow a^3+ab^2+ac^2-a^2b-abc-ca^2+ba^2+b^3+bc^2-b^2a-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2=0\)

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

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

\(\Rightarrowđpcm\)