Áp dụng hằng đẳng thức mở rộng :
\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Rightarrow A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(\Leftrightarrow A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2}\)
\(\Leftrightarrow A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2\left(a^2+b^2+c^2-ab-bc-ac\right)}\)
\(\Leftrightarrow A=\dfrac{a+b+c}{2}\)
Thay \(a+b+c=4\) vào M ta có :
\(A=\dfrac{4}{2}=2\)
Vậy A = 2
