$\large \frac{1}{(1-x)(1-y)(1-z)}+\frac{1}{(1+x)(1+y)(1+z)}$
For $-1<x,y,z<1$, let $t$ denote the minimum value of the above expression and $u$ be the sum of $x,y,z$ when the equality holds, find $(3t)^{u}+4$.

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