a³ + b³ + c³ = 3abc

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

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

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

=> (a + b + c)³ = 0

=> a + b + c = 0 (đpcm)

           a + b + c = 0

\(\Leftrightarrow\)(a + b + c)³ = 0

\(\Leftrightarrow\)a³ + b³ + c³ + 3(a²b + ab² + a²c + ac² + b²c + bc² + 2abc) = 0

\(\Leftrightarrow\)a³ + b³ + c³ + 3[ab(a + b + c) + ac(a + b + c) + bc(b + c)] = 0

\(\Leftrightarrow\)a³ + b³ + c³ + 3bc(b + c) = 0

\(\Leftrightarrow\)a³ + b³ + c³ = -3bc(b + c)

mà a + b + c = 0

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

Ta có:   a³ + b³ + c³ = -3bc(b + c)

\(\Leftrightarrow\)a³ + b³ + c³ = -3bc(-a)

\(\Leftrightarrow\)a³ + b³ + c³ = 3abc    (đpcm)

Ta có \(a+b+c=0\)

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

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

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

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

Mà \(a+b=-c\)

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

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

a + b + c = 0 => a = -b - c

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

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

= \(-3b^2c-3cb^2\)

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

= \(3abc=VP\)

Áp dụng hằng đẳng thức:\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

       Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}\left(1\right)}\)

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

           Lấy (1) thay vào (2) ta được:

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

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

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

Ta có : a + b + c = 0

=> ( a + b + c )³ = 0

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

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

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

MÀ a + b + c = 0

=> a³ + b³ + c³ = 3abc (đpcm)

Hok Tốt!!!!

ta có :a+b+c=0 \(\Rightarrow\) a + b  =\(-\)c (1)   \(\Rightarrow\) (a+b)\(^3\) = ( -c )\(^3\)\(\Rightarrow\) \(a^{3^{ }}+3a^2+3ab^2+b_{ }^2\) = -c^3

\(\Leftrightarrow\)a^3 + b^3 + c^3 + 3ab(a+b) =0   từ (1) \(\Rightarrow\)a^3 + b^3 + c^3 =\([3ab\left(-c\right)]=0\) 

\(\Rightarrow a^{3^{ }}+b^3+c^3-3abc=0\) \(\Leftrightarrow a^3+b^3+c^3=3abc\) \(\Rightarrow\) ĐPCM

Lời giải:

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

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

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

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

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

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

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

Với $a^2+b^2+c^2-ab-bc-ac=0$

$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$

$\Rightarrow (a-b)^2=(b-c)^2=(c-a)^2=0$

$\Leftrightarrow a=b=c$

Do đó ta có đpcm.

Ta có : a³ + b³ + c³ = 3abc

=> a³ + b³ + c³ -3abc = 0

=> ( a + b )³ - 3ab( a - b ) + c³ -3abc = 0

=> ( a + b )³ + c³- [(3ab( a +b) + 3abc] =0

=> ( 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 - bc - ac ) = 0

=> a + b + c = 0 ( đpcm ) hoặc có thể là a = b = c ( đpcm ) nhé.

thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có : 

a^3+b^3+c^3-3abc=0 

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc=0 

câu 2:<=>[(a+b)^3 +c^3] -3ab.(a+b+c)=0 

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

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)... 

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0 

luôn đúng do a+b+c=0

câu 1:(a+b+c)^3=((a+b)+c)^3=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c))
=a^3+b^3+c3^+3(a+b)(ab+ac+bc+c2)
=a^3+b^3+c^3+3(a+b)(a+c)(b+c)

CHÚC BẠN HỌC TỐT^^

Có : a + b + c  = 0 

=> a + b = -c

Vậy a³ + b³ + c³ = (a + b)³ - 3ab(a + b) + c³

= (-c)³ + 3abc + c³

= 3abc (=VP)

\(a+b+c=0\Rightarrow a+b=-c\)

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

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

mà a+b= -c (cmt )

nên \(a^3+b^3+c^3=3abc\left(đpcm\right)\)

\(a+b+c=0\Rightarrow c=-a-b\)

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

\(=-3a^2b-3ab^2=3ab\left(-a-b\right)=3abc\) (đpcm)

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

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

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

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

Theo đề ta có: 
a+b+c=0 => c=-(a+b) (1) 
Thay (1) vao a^3+b^3+c^3 ta có: 
a³+b³+[-(a+b)]³=3ab[-(a+b)] 
<=>a³+b³-(a+b)=-3ab(a+b) 
<=> a3+ b3- a3 -3a2b- 3ab2- b3= -3a2b- 3ab2 
<=> 0= 0 
vậy ta có đpcm.

b) có a+b+c = 0 
=> a²+b²+c²+2(ab+bc+ac) = 0
mà a²+b²+c² = 2
=> ab+bc+ac = -1
=> a²b²+b²c²+a²c² + 2ab²c+2a²bc+2abc² = 1
=>a²b²+b²c²+a²c² + 2abc(b+a+c) = 1
=>a²b²+b²c²+a²c² = 1
Ta bìn phong cái a²+b²+c² len 
đk là
a⁴+b⁴+c⁴ + 2a²b²+2a²c²+2b²c²=4
=> a⁴+b⁴+c⁴ + 2(a²b²+a²c²+b²c²) = 4
mà ở trên là a²b²+b²c²+a²c² = 1
=> a⁴+b⁴+c⁴ +1 =4
a⁴+b⁴+c⁴ = 3 D

k giùm nha!!!