\(a^3-b^3+c^3+3abc\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)+c^3+3abc\)
\(=\left\lbrack\left(a-b\right)+c\right\rbrack\left\lbrack\left(a-b\right)^2-c\left(a-b\right)+c^2\right\rbrack+3ab\left(a-b+c\right)\)
\(=\left(a-b\right)^2-c\left(a-b\right)+c^2+3ab\)
\(=a^2-2ab+b^2-ac+bc+c^2+3ab=a^2+b^2+c^2+ab-ac+bc\)
\(\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2-2ac+c^2\)
\(=2a^2+2b^2+2c^2+2ab+2bc-2ac\)
\(=2\left(a^2+b^2+c^2+ab+bc-ac\right)\)
Ta có: \(M=\frac{a^3-b^3+c^3+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\frac{a^2+b^2+c^2+ab-ac+bc}{2\left(a^2+b^2+c^2+ab-ac+bc\right)}=\frac12\)
