\(a^3+b^3+c^3=3abc\)(1)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]=0\)
\(\Rightarrow a+b+c=0\)(2)
Từ 1 và 2
\(\Rightarrow a=b=c\)
#Đức Lộc#
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc+3a^2b+3ab^2-3a^2b-3ab^2=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3abc-3a^2b-3ab^2=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ac-bc-ab=0\left(2\right)\end{cases}}\)
Từ (1) ta có đpcm
\(\left(2\right)\Leftrightarrow2a^2+2b^2+2c^2-2ac-2bc-2ab=0\)
\(\Leftrightarrow a^2-2ac+c^2+b^2-2bc+c^2+a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-c\right)^2+\left(b-c\right)^2+\left(a-b\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-c=0\\b-c=0\\a-b=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=c\\b=c\\a=b\end{cases}\Leftrightarrow a=b=c}}\)( đpcm )
Vậy ta có đpcm
