C1: Ta có: \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\) (1)
Ta có: \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^3=0^3\)
\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\) (2)
Thay (1) vào (2) ta có:
\(a^3+b^3+c^3+3.\left(-a\right).\left(-b\right).\left(-c\right)=0\)
\(a^3+b^3+c^3-3abc=0\)
\(a^3+b^3+c^3=3abc\)
đpcm
C2: \(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(a^3+3a^2+3ab^2+b^2=-c^3\)
\(a^3+b^3+c^3+3ab\left(a+b\right)=0\)
Ta có: \(a+b=-c\)
\(\Rightarrow\)\(a^3+b^3+c^3+3ab\left(-c\right)=0\)
\(a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
đpcm
Ta có:\(a+b+c=0\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3=-c^3\Leftrightarrow a^3+3ab\left(a+b\right)+b^3=-c^3\)
\(\Leftrightarrow a^3-3abc+b^3=-c^3\Leftrightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)
Ta có: a + b + c = 0 => a +b = -c
\(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)\)\(+c^3\)
\(=-c^3-3ab\left(-c\right)+c^3\)
\(=3abc\)(đpcm)
Vậy nếu a + b + c =0 thì \(a^3+b^3+c^3=3abc\)
thế méo nào \(^{a^3+b^3+c^3=\left(a+b+c\right)^3}\) được
chỉ co nhan moi ghep dc thoi
lua dao
