17. If $a, b$ and $c$ are positive real numbers such that
$$
a b+a+b=b c+b+c=c a+c+a=35 \text {, }
$$
find the value of $(a+1)(b+1)(c+1)$.
17. Answer: 216
Adding 1 to both sides of the given equation $a b+a+b=35$, we obtain
$$
(a+1)(b+1)=36 \text {. }
$$

Likewise, adding 1 to the other two given equations gives
$$
(b+1)(c+1)=36 \text { and }(c+1)(a+1)=36 \text {. }
$$

Now multiplying the three resulting equations above leads to
$$
[(a+1)(b+1)(c+1)]^{2}=36^{3}=6^{6} .
$$

It follows that $(a+1)(b+1)(c+1)=6^{3}=216$.