If $abc+1=0$, verify that' If $abc+1=0$, prove that:
$\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}} +\frac {1}{1-c-a^{-1}}=1$.
My Attempt:
$abc+1=0$
$abc=-1$.
Now,
$$L.H.S=\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}}+\frac {1}{1-c-a^{-1}}$$
$$=\frac{b}{b-ab-1} + \frac {1}{1-b-c^{-1}} + \frac {c^{-1}}{c^{-1}-1-(ca)^{-1}}$$
$$=\frac{b}{b+c^{-1} -1} +\frac{1}{1-b-c^{-1}} +\frac{c^{-1}}{c^{-1}-1+b}$$
What should I do next?

$$\dfrac1{1-a-b^{-1}}=\dfrac{bc}{bc-abc-c}=\dfrac{bc}{bc+1-c}$$
$$\dfrac1{1-b-c^{-1}}=\dfrac{-c}{-c+bc+1}$$
As $-abc=1,-a^{-1}=bc$
$$\dfrac1{1-c-a^{-1}}=\dfrac1{1-c+bc}$$
