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

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

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

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

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

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

Từ a + b + c = 0 suy ra được : \(c=-\left(a+b\right)\Rightarrow c^3=-\left[a^3+b^3+3ab\left(a+b\right)\right]\Rightarrow a^3+b^3+c^3=-3ab\left(-c\right)=3abc\)

Vậy : \(a^3+b^3+c^3-3abc=0\)

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:

     \(a^3+b^3+c^3-3abc\)

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

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

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

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

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

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

Mà \(a+b+c=0\)

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

Vậy: nếu \(a+b+c=0\) thì \(a^3+b^3+c^3-3abc=0\)

Chúc bạn học tốt và tíck cho mìk vs nha bùi thị thu hương!

bùi thị thu hương khó ứa

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

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

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

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

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

Vì a,b,c > 0 nên a+b+c > 0

Do đó : \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow}a=b=c\)

1) có: a^3 + b^3 + c^3 - 3abc = 0
((a + b)3 + c^3( - 3ab(a + b) - 3abc = 0
<=>(a + b + c)((a + b)2 - (a + b).c + c2( - 3ab(a + b + c) = 0
<=>(a + b + c) (a2 + b2 + c2- ac - bc - ab( = 0

Từ đây cho nhận xét:
+ Nếu a + b + c = 0 có a3 + b3 + c3 = 3abc (I)
a + b + c = 0 
+ Nếu a^3 + b^3 + c^3 = 3abc thì 
a = b = c

ak thôi CHTT có rồi,mn khỏi phải giải

a ) \(a^3+b^3+c^3=3abc\)

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

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

Nếu : \(a+b+c=0\) thì đẳng thức trên đúng .

\(\Rightarrowđpcm\)

b ) \(a+b+c+d=0\)

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

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

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

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(cb-cd\right)\left(đpcm\right)\)

Chúc bạn học tốt !!!

a ) a^3+b^3+c^3=3abca3+b3+c3=3abc

\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0⇔(a+b)3+c3−3ab(a+b)−3abc=0

\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0⇔(a+b+c)(a2+b2+c2−ab−bc−ac)=0

Nếu : a+b+c=0a+b+c=0 thì đẳng thức trên đúng .(đpcm)

b ) a+b+c+d=0a+b+c+d=0

\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3⇒a+b=−(c+d)⇔(a+b)3=−(c+d)3

\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)⇔a3+b3+c3+d3=−3ab(a+b)−3cd(c+d)

\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)⇔a3+b3+c3+d3=3ab(c+d)−3cd(c+d)

\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(cb-cd\right)\left(đpcm\right)⇔a3+b3+c3+d3=3(c+d)(cb−cd)(đpcm)