Từ \(a^3+b^3+c^3=3abc\)\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\left[\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
*)Xét \(a=b=c\). Khi đó \(\frac{a^{2011}}{b^{2011}}+\frac{b^{2011}}{c^{2011}}+\frac{c^{2011}}{a^{2011}}=1+1+1=3\)
*)Xét \(a+b+c=0\Rightarrow\)\(\left\{\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\). Khi đó \(\frac{a^{2011}}{b^{2011}}+\frac{b^{2011}}{c^{2011}}+\frac{c^{2011}}{a^{2011}}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
bạn có cần cách giải ko, mình r ết quả = 3 đó