Sửa đề: \(a^3+b^3+c^3=3abc\) và a+b+c<>0
Ta có: \(a^3+b^3+c^3=3abc\)
=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3bac=0\)
=>\(\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
=>\(a^2+b^2+c^2-ab-ac-bc=0\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
=>\(\begin{cases}a-b=0\\ b-c=0\\ a-c=0\end{cases}\Rightarrow a=b=c\)
Ta có: \(N=\frac{a^2+b^2+c^2}{a+b+c}\)
\(=\frac{a^2+a^2+a^2}{a+a+a}=\frac{3a^2}{3a}=a\)
