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.

Since I prefer working with exponents rather than logarithms, here is how you could tackle it that way, using your quite possibly better intuition at handling powers:
$$x + y + z = 8$$
$$u + v + w = 13$$
With:
$$ a^x = b\,,\, b^y = c\,,\, c^z = a $$
and
$$ b^u = a\,,\, c^v = b\,,\, a^w = c $$
From this you can see that:
$$ a^{x\,u} = b^u = a \; \Rightarrow \; x u = 1$$
So the second equation is actually:
$$ 1/x + 1/y + 1/z = 13 $$
If you add the fractions:
$$ yz + xz + xy = 13 \, x\,y\,z $$
Since:
$$ (1+x)(1+y)(1+z) = xyz + xz + yz + xy + x + y + z + 1  $$
We need to figure out what is $xyz$. But:
$$ a^{x\,y\,z} = b ^{y\,z} = c^z = a \; \Rightarrow \; x y z = 1 $$ 
Now we have all the information we need and the result is $23$.
