Ta có :
\(a^3+b^3+c^3-3abc\)
\(=a^3+b^3+c^3+3a^2b+3ab^2-3a^2b-3ab^2-3abc\)
\(=\left[\left(a^3+3a^2b+3ab^2+b^3\right)+c^3\right]-\left(3a^2b+3ab^2+3abc\right)\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left\{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\right\}-3ab\left(a+b+c\right)\)
\(=\left[\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Rightarrow P=\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=a+b+c\)
\(=2014\)
Vậy P = 2014
