If $a, b, c$ are positive real numbers such that $abc=1$ prove that $\frac{a^3}{(a-b)(a-c)} + \frac{b^3}{(b-a)(b-c)} + \frac{c^3}{(c-b)(c-a)} ≥ 3$
How find this value $\frac{a^3}{(a-p)(a-q)}+\frac{b^3}{(b-p)(b-q)}+\frac{c^3}{(c-p)(c-q)}$