Let $\displaystyle f(x|\alpha)=\frac{\Gamma(2\alpha)}{\Gamma(\alpha)^ 2}[x(1-x)]^{\alpha -1}$

Then the joint pdf= $\displaystyle \prod\limits_{i=1}^n f(x|\alpha)$

= $\displaystyle (\frac{\Gamma(2\alpha)}{\Gamma(\alpha)^2})^n \prod\limits_{i=1}^n[x_i(1-x_i)]^{\alpha -1} $

by factorization theorem, $\displaystyle g(t,\alpha)h(x)$

$\displaystyle t=\prod\limits_{i=1}^n[x_i(1-x_i)]$

$\displaystyle g(t,\alpha)=(\frac{\Gamma(2\alpha)}{\Gamma(\alpha) ^2})^n t^{\alpha -1}$

$\displaystyle h(x)=1$

Can I conclude that $\displaystyle T(X)=\prod\limits_{i=1}^n[X_i(1-X_i)]$ is a sufficient statistic?