Question

 a/ cho a+b+c=0 . cm  a^3  + b^3 + c^3 = 3abc

Answer 1

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:

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

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

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

Answer 2

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!!!!

Answer 3

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