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.

We have the following:
$$(log_ab+1)(log_bc+1)(log_ca+1)=(\frac {logb}{loga}+1)(\frac {logc}{logb}+1)(\frac {loga}{logc}+1)$$
Expanding yields:
$$(\frac {logc}{loga}+\frac {logb}{loga}+\frac {logc}{logb}+1)(\frac {loga}{logc}+1)$$
$$=(\frac {loga}{loga}+\frac {logc}{loga}+\frac {logb}{logc}+\frac {logb}{loga} +\frac {loga}{logb}+\frac {logc}{logb}+\frac {loga}{logc}+1)$$
$$=(log_aa+log_ac+log_cb+log_ab+log_ba+log_bc+log_ca+1)$$
$$=(log_ab+log_bc+log_ca)+(log_ba+log_cb+log_ac)+2$$
Thus finally we see, via your initial condition:
$$(log_ab+1)(log_bc+1)(log_ca+1)=8+13+2=23$$
