Given $g^a, g^b, g^c, g^{1/b}$, is it hard to distinguish $e(g, g)^{abc}$ from a random value? where $g$ is a group element in bilinear group $\mathbb{G}$. I understand it is very similar to the conventional DBDH problem, but $g^{1/b}$ is also known, possibly making it easier? Does anyone know the answer or suggest some material for reference? Thanks.
Your problem seems to be at least as hard as the 2-weak Bilinear Diffie-Hellman Inversion Problem (2-wBDHI problem):


  Given $g, g^x, g^{x^2}, g^y \in \mathbb G$, and $T \in \mathbb G_T$ to determine whether or not $T = e(g,g)^{x^3 y}$.


Proof: We first need to define an equivalent version of your problem, where we take some generator $h$ so $g = h^b$. Your original problem is to take input $(g,g^a, g^b, g^c, g^{1/b}, Q)$ and decide whether $Q = e(g,g)^{abc}$ or not. After substituting $g = h^b$, we have that the new problem is to take $(h^b, h^{ab}, h^{b^2}, h^{bc}, h, Q)$ and decide if $Q = e(h,h)^{b^3 ac}$.

So, assume you have a solver $S$ for the new problem.  Then we can solve the 2-wDBDHI problem as follows: 


Input to 2-wDBDHI is a tuple $(g,g^x,g^{x^2},g^y,T)$. 
Sample a random element $z \in \mathbb Z_q^*$.
Call $S$ with input $(g^x,g^y,g^{x^2},g^{x^2 z},g,T^z)$, which will output Yes when $T^z= e(g,g)^{x^3 y z}$, and No otherwise.
Output the result from the last step.


Recall that solver $S$ takes input $(h^b, h^{ab}, h^{b^2}, h^{bc}, h, Q)$ and determines whether the following equation holds:
\begin{align}Q = e(h,h)^{b^3 ac}\end{align}
Take the substitutions $h=g$, $a = y/x$, $b = x$, $c = xz$, and $Q = T^z$, so:
\begin{align}T^z = e(g,g)^{x^3 (y/x) xz} = e(g,g)^{x^3 yz}\end{align}
It is clear that when this equation holds, then $T = e(g,g)^{x^3 y}$ holds too, and vice versa, so this algorithm is a solver for the 2-wBDHI problem. 

According to the ECRYPT II report - D.MAYA.6
Final Report on Main Computational Assumptions in
Cryptography, the best known algorithm for 2-wBDHI is to solve DLP in $\mathbb G$, so you can say that your problem is hard. 

Note: My previous answer didn't answer your question, since the strong-DDH is not hard when using pairing groups... I was merely stating that your problem is at least as hard as an easy problem (duh!)