8) The number $\frac{1}{(3+\sqrt{2})(2+\sqrt{3})(3-\sqrt{2})(2-\sqrt{3})}$ is:

(A) Integer (B) positive rational, but not an integer

(C) negative rational, but not an integer

(D) positive irrational (E) negative irrational.
8) The answer is $(\mathbf{B})$. We use the identity $(a+b)(a-b)=a^{2}-b^{2}$ twice, once with $a=3, b=\sqrt{2}$ and the other with $a=2, b=\sqrt{3}$: we have that

$$
(3+\sqrt{2})(3-\sqrt{2})=3^{2}-(\sqrt{2})^{2}=9-2=7
$$

and

$$
(2+\sqrt{3})(2-\sqrt{3})=2^{2}-(\sqrt{3})^{2}=4-3=1
$$

thus $\frac{1}{(3+\sqrt{2})(2+\sqrt{3})(3-\sqrt{2})(2-\sqrt{3})}=\frac{1}{(9-2)(4-3)}=\frac{1}{7}$ and the original fraction simplifies to $\frac{1}{7}$, which is a positive rational (non-integer) number.