If $B(\theta,2\theta)$, how do you show that $\prod X_1(1-X_1)^2$ is a sufficient statistic for $\theta$? A random sample $X_1,\ldots,X_n$ is taken from a distribution with probability density function which is $B(\theta,2\theta)$. Show that $\prod X_i(1-X_i)^2$ is a sufficient statistic for $\theta$.
Below is the beta distribution with the parameters referred to:
$$f_X(x;\theta)=\frac{\Gamma(3\theta)}{\Gamma(\theta)(\Gamma(2\theta)}x^{\theta-1}(1-x)^{2\theta-1}$$
To be honest, this question is beyond me. I don't know how to start. 
I have been reading up about sufficient statistics, and using factorisation theorem, but I don't know if this is the right approach and I don't really know where to begin with this.
Do I need to work out a joint pdf? Do I need to know the likelihood function? Is factorisation theorem even the right thing to use?
Any help (even if it is just directing me to a source for "sufficient statistics for dummies") is much appreciated! 
 A: Indeed you can use the factorization theorem and show that 
$$
\prod_{i=1}^n f(x_i ; \theta)= \frac{1}{B ^n( \theta, 2\theta)} \left( \prod_{i=1}^n x_i ( 1 - x_i) ^ 2 \right) ^ {\theta}  \left(\prod_{i=1}^nx_i(1-x_i)\right)^{-1}
$$
thus, considering
$$T(x) =\prod_{i=1}^n x_i ( 1 - x_i)^2 $$ one sees that  $$g_{\theta}(T(x)) =  \frac{1}{B ^ n( \theta, 2\theta)} \left( \prod_{i=1}^n x_i ( 1 - x_i) ^ 2 \right) ^ {\theta}$$ so $T(x) $ is the MSS. 
EDIT:
You should start with the likelihood function (not the log-likelihood), and then  "separate" everything known (the sample or its function) from everything unknown (depends on $\theta$). Then find the function $g$ where $\theta$ depends on some function of the sample - which will be the MSS. Namely, start by reformulating the likelihood itself
\begin{align}
L(\theta; x_1,...,x_n) &= \prod_{i=1}^n ( B (\theta, 2\theta) ) ^{-1}
 x_i^{\theta} x_i ^{-1} ( ( 1 - x_i) ^2)^{\theta}  (1-x_i) ^{-1} \\
&=\prod_{i=1}^n ( B (\theta, 2\theta) ) ^{-1}
   (x_i( 1 - x_i ) ^ 2 )^{\theta} ((1-x_i)x_i) ^{-1} \\
 & =   ( B (\theta, 2\theta) ) ^{-n}
   \left(\prod_{i=1}^n x_i( 1 - x_i ) ^ 2 \right)^{\theta}
 \prod_{i=1}^n\left((1-x_i)x_i \right) ^{-1}. 
\end{align}
Although the MLE is indeed a function of the MSS, however it is "overkill" to derive the MLE in order to find the MSS. 
