Find a probability density function of three variables such that

1. It is symmetric
2. The marginal density of the remaining two are separable aka pair-wise independent
3. It is not separable aka independent among the three variables

Let $F(x,y,z)=( 2 x y z + x + y + z + \frac{1}{2} ) \frac{4}{9}$ the pdf of three $[0,1$] interval random variables. $F$ can not be separable. If $F$ would be separable then $F(x,y,z) = H(x)H(y)H(z)$ . Put $y = z = x$ then

$\begin{aligned}H(x)= \left(\left( 2 x^3 + 3 x + \frac{1}{2}\right) \frac{4}{9}\right)^{\frac{1}{3}}\end{aligned}$

but $F(x,y,z)= H(x)H(y)H(z)$ is not true.

The marginal pdf of two variables is $G(x,y) = (x y + x + y + 1)\frac{4}{9} = (x + 1 ) ( y + 1) \frac{4}{9}$

And marginal pdf of one variable is $H(x)= (x + 1)\frac{2}{3}$ and $G(x,y) = H(x)H(y)$