Have You Practice Your Algebra Today?

It is easy to see that $b=a+1$ and $c=a+2$ . Then using that and the fact that $abc=3$ $\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$ $=\frac{(a^{2}+b^{2}+c^{2})-(ab+bc+ac)}{abc}$ $=\frac{(a^{2}+(a+1)^{2}+(a+2)^{2})-(a(a+1)+(a+1)(a+1)+a(a+2))}{3}$ $=\frac{3}{3}$ $=1$