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

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

\(\Rightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b+c\right)=0\)

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

\(\left(a+b+c\right)\left(a^2-ab+b^2-bc+c^2-ca\right)=0\)\(Màa,b,c\ne0\Rightarrow a^2-ab+b^2-bc+c^2-ca=0\Rightarrow a\left(a-b\right)+b\left(b-c\right)+c\left(c-a\right)=0\)

\(a,b,c\ne0\Rightarrow a-b=0;b-c=0;c-a=0\Rightarrow a=b=c\)

Ta có : a + b + c = 0 => a = -(b + c)

Nên a³ + b³ + c³ - 3abc

= [-(b + c)]³ + b³ + c³ - 3abc

= -(b³ + 3b²c + 3bc² + c³) + b³ + c³ ​- 3abc

= -b³ - 3b²c - 3bc² - c³ + b³ + c³ ​- 3abc

= -3bc(a + b + c) 

Mà a + b + c = 0 

=> 3bc(a + b + c) = 0

Vậy a³ + b³ + c³ - 3abc = 0 (đpcm)

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

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

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

b,

Ta có:

\(\left(a+b+c\right)^3=0\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow a^3+b^3+c^3-3.\left(-c\right)\left(-a\right)\left(-b\right)=0\)

Ta có a^3 + b^3 + c^3 = (a+b+c). (a^2+b^2+c^2-a.b-b.c-a.c)+3abc= 3abc

                                = (a+b+c)(a^2+b^2+c^2-a.b-b.c-a.c)=0

Ta Thấy a,b,c là số dương nên a+b+c khác 0 suy ra ( a^2+b^2+c^2-a.b-b.c-a.c)=0 Nên a=b=c

- k Mình Nhé 

Ta có: a³ + b³ + c³ = 3abc

<=> a³ + b³ + c³ − 3abc = 0

<=> (a + b + c) (a² + b² + c² − ab − bc − ca) = 0

<=> a² + b² + c² − ab − bc − ca = 0 (do a + b + c > 0)

<=> 1/2(2a² + 2b² + 2c² − 2ab − 2bc − 2ca) = 0

<=> a² - 2ab + b² + b² - 2bc + c² + c² - 2ac + a² = 0 

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

<=> a − b = b − c = c − a = 0

<=> a = b = c 

`(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`

Ta có a,b,c dương nên ta áp dụng Bđt Cô-si ta có:

\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)

Dấu = khi \(a=b=c\)

Đpcm

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

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

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

\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right]=0\)

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

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

\(\Rightarrow\left(a+b+c\right)\frac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)=0\)

\(\Rightarrow\left(a+b+c\right)\frac{1}{2}\left(a^2-2ab+b^2\right)\left(b^2-2bc+c^2\right)\left(c^2-2ac+a^2\right)=0\)

\(\Rightarrow\left(a+b+c\right)\frac{1}{2}\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0.\)

vì \(\left(a-b\right)^2\ge0\)

\(\left(b-c\right)^2\ge0\)

\(\left(c-a\right)^2\ge0\)

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

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

\(\Rightarrow a=b=c\left(dpcm\right)\)