Ta có:

 \(a^2+b^2+c^2=ab+bc+ca\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\ \Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(\left(a-b\right)^2,\left(b-c\right)^2,\left(c-a\right)^2\ge0\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Rightarrow\left(a-b\right)^2=\left(b-c\right)^2=\left(c-a\right)^2=0\\ \Leftrightarrow a=b=c\)

Lại có: \(a+b+c=3\Rightarrow a=b=c=1\)

\(\Rightarrow M=1^{2016}+1^{2015}+1^{2020}=1+1+1=3\)

dễ mà!  mọi người cứ làm quá lên

không trả lời được mà dễ

Ta có :

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

Do \(a+b=a^3+b^3\)

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

\(\Rightarrow a^2-ab+b^2=1\)

Mà \(a^2=b^2=a+b\) ,ta có :

\(a+b-ab=1\)

\(\Rightarrow a+b-ab-1=0\)

\(\Rightarrow\left(a-1\right)-\left(ab-b\right)=0\)

\(\Rightarrow\left(a-1\right)-b\left(a-1\right)=0\)

\(\Rightarrow\left(a-1\right)\left(1-b\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}a-1=0\\1-b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

Thay vaò biểu thức ,có :

\(1^{2015}+1^{2015}=1+1=2\)

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

Xét \(a+b+c\ne0\) ta có:\(\frac{a-b+c}{b}=\frac{b-c+a}{c}=\frac{c-a+b}{a}=\frac{a-b+c+b-c+a+c-a+b}{a+b+c}=1\)

\(\Rightarrow\hept{\begin{cases}a-b+c=b\\b-c+a=c\\c-a+b=a\end{cases}}\Rightarrow\hept{\begin{cases}a+c=2b\\a+b=2c\\b+c=2a\end{cases}}\Rightarrow M=\frac{2a.2b.2c}{abc}=8\)