Từ \(a+b+c=0\Leftrightarrow a+b=-c\)

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

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

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

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

ta co :a + b+c=0

=>(a+b+c)^3= 0

<=>  a^3 + b^3 + c^3 + 3a^2b+3a^2c + 3b^2a+3b^2c + 3c^2a+3c^2b + 6abc =0

<=>(a^3+b^3+c^3) + (3a^2b+3a^2c+3abc ) +(3b^2a+3b^c +3abc) +(3c^2a+3c^b +3abc )  - 3abc=0

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

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

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

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

=>a^3+b^3+c^3 =3abc (dpcm)

Ta co

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

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

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

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

Ma a+b+c=3

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

=>\(a^3+b^3+c^3=3abc\)(\(ĐPCM\))

a+b+c=0 nhe minh ghi nham =3

Ta có a³ + b³ + c³ = 3abc

<=> (a + b)³  - 3ab(a + b) + c³ = 3abc

<=> (a + b + c)[(a + b)² - (a + b)c + c²] - 3ab(a + b + c) = 0

<=> (a + b + c)(a² + 2ab + b² - ac - bc + c² - 3ab) = 0 

<=> (a + b + c)(a² + b² + c² - ab - ac - bc) = 0

<=> \(\orbr{\begin{cases}a+b+c=0\left(\text{tmđk}\right)\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)

Khi a² + b² + c² - ab - ac - bc = 0 

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

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

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

<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(\text{loại}\right)\)

Vậy a + b + c = 0

Ta dùng cách chứng minh ngược :

Nếu \(a=b=c\) thì \(a^3=b^3=c^3=abc\)

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

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