Question

Answer 1

Ta có \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\\ =\dfrac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\\ =\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

\(\Leftrightarrow M=\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\cdot2022=1011\)