Power of 1

$\dfrac{a^3}{(a-b)(a-c)} + \dfrac{b^3}{(b-a)(b-c)} + \dfrac{c^3}{(c-a)(c-b)}\\ = \dfrac{a^3}{(a-b)(a-c)} - \dfrac{b^3}{(a-b)(b-c)} + \dfrac{c^3}{(a-c)(b-c)} \\ = \dfrac{a^3(b-c) - b^3(a-c) + c^3(a-b)}{(a-b)(a-c)(b-c)} \\ = \dfrac{a^3b - a^3c - b^3a + b^3c + c^3a - c^3b}{(a-b)(a-c)(b-c)} \\ = \dfrac{a^3b - b^3a - a^3c + b^3c+c^3a - c^3b)}{(a-b)(a-c)(b-c)}\\ = \dfrac{ab(a^2 - b^2) -c(a^3 - b^3) + c^3(a-b)}{(a-b)(a-c)(b-c)} \\ = \dfrac{ab(a+b)(a-b) - c (a-b)(a^2 + ab +b^2) + c^3(a-b)}{(a-b)(a-c)(b-c)} \\ = (a-b)\dfrac{ab(a+b) - c(a^2 + ab + b^2) + c^3}{(a-b)(a-c)(b-c)} \\ = \dfrac{a^2b + b^2a - a^2c - abc - b^2c + c^3}{(a-c)(b-c)} \\ = \dfrac{a^2b - a^2c - b^2c + c^3 + b^2a - abc}{(a-c)(b-c)}\\ \\ = \dfrac{a^2(b-c) -c(b^2 - c^2) + ab(b-c)}{(a-c)(b-c)}\\ = \dfrac{a^2(b-c) - c(b+c)(b-c) + ab(b-c)}{(a-c)(b-c)}\\ = \dfrac{a^2 - c(b+c) + ab}{(a-c)} \\ = \dfrac{a^2 - bc - c^2 + ab}{(a-c)} \\ = \dfrac{a^2 - c^2 - bc + ab}{(a-c)} \\ = \dfrac{(a+c)(a-c) + b(a-c)}{(a-c)} \\ = a+b+c \\ \boxed{\therefore \dfrac{a^3}{(a-b)(a-c)} + \dfrac{b^3}{(b-a)(b-c)} + \dfrac{c^3}{(c-a)(c-b)} = a+b+c} \\ \text{Using AM-GM inequality, and noting a, b, c are distinct} \\ \dfrac{a+b+c}{3} > \sqrt[3]{abc} \\ abc = 1 \\ \implies \dfrac{a+b+c}{3} > 1 \\ \implies a+b+c > 3 \\ \boxed{\therefore \dfrac{a^3}{(a-b)(a-c)} + \dfrac{b^3}{(b-a)(b-c)} + \dfrac{c^3}{(c-a)(c-b)} = a+b+c > 3}$