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\[\begin{eqnarray}xyz + 3x + 3yz&=&106 \\ xyz + 3z + 3xy&=&79 \\ xyz+ 3y + 3xz&=&82 \\ (x+1)(y+1)(z+1) &=& \ ? \end{eqnarray} \]
See Part 1 and Part 2.
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