Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that 
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and 
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of 
$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1) ?$$
I tried to convert the entire thing to fractional logs and multiply the expression and add the two equations but it did not help.

Notice that
$$
\log_uv\log_vw=\dfrac{\log v}{\log u}\cdot\dfrac{\log w}{\log v}=\dfrac{\log w}{\log u}=\log_uw \quad \forall u,v,w>0
$$
therefore
\begin{eqnarray}
(\log_ab+1)(\log_bc+1)(\log_ca+1)&=&(\log_ab\log_bc+\log_ab+\log_bc+1)(\log_ca+1)\\
&=&(\log_ac+\log_ab+\log_bc+1)(\log_ca+1)\\
&=&\log_ac\log_ca+\log_ac+\log_ab\log_ca+\log_ab+\log_bc\log_ca+\log_bc+\log_ca+1\\
&=&\log_aa+\log_ac+\log_cb+\log_ab+\log_ba+\log_bc+\log_ca+1\\
&=&(1+\log_aa)+(\log_ab+\log_bc+\log_ca)+(\log_ba+\log_cb+\log_ac)\\
&=&1+1+8+13=23.
\end{eqnarray}
