Ta có A=\(\dfrac{a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{a^3\left(b-c\right)+b^3c-c^3b-a\left(b^3-c^3\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{a^2\left(b-c\right)+bc\left(b^2-c^2\right)-a\left(b-c\right)\left(b^2+bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)