24. The numbers $x, y$ and $z$ are given by $x=\sqrt{12-3 \sqrt{7}}-\sqrt{12+3 \sqrt{7}}, y=\sqrt{7-4 \sqrt{3}}-$ $\sqrt{7+4 \sqrt{3}}$ and $z=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}$.
What is the value of $x y z$ ?
A 1
B -6
C -8
D 18
E 12
Solution
E
COMMENTARY
A natural first thought on reading this question is that working out the product $x y z$, without the use of a calculator, will be horribly complicated, and that, given the time constraints of the SMC, it would be sensible to skip this question.

But then you might realize that, because this is an SMC question, there could well be a smarter way to tackle this question.

Because the expressions for $x, y$ and $z$ involve lots of square roots, a smart method might be to work out $x^{2}, y^{2}$ and $z^{2}$ separately and hence find $x^{2} y^{2} z^{2}$, which is equal to $(x y z)^{2}$. It is, perhaps, surprising that this approach works out well.

We have
$$
\begin{array}{l}
x^{2}=(\sqrt{12-3 \sqrt{7}}-\sqrt{12+3 \sqrt{7}})^{2} \\
=(\sqrt{12-3 \sqrt{7}})^{2}-2(\sqrt{12-3 \sqrt{7}})(\sqrt{12+3 \sqrt{7}})+(\sqrt{12+3 \sqrt{7}})^{2} \\
=(\sqrt{12-3 \sqrt{7}})^{2}-2 \sqrt{((12-3 \sqrt{7})(12+3 \sqrt{7}))}+(\sqrt{12+3 \sqrt{7}})^{2} \\
=(\sqrt{12-3 \sqrt{7}})^{2}-2\left(\sqrt{12^{2}-(3 \sqrt{7})^{2}}\right)+(\sqrt{12+3 \sqrt{7}})^{2} \\
=(12-3 \sqrt{7})-2 \sqrt{144-63}+(12+3 \sqrt{7}) \\
=(12-3 \sqrt{7})-2 \sqrt{81}+(12+3 \sqrt{7}) \\
=24-2 \times 9 \\
=6 \text {. } \\
\end{array}
$$

We leave it as an exercise to check that, similarly, $y^{2}=12$ and $z^{2}=2$. It follows that $(x y z)^{2}=x^{2} y^{2} z^{2}=6 \times 12 \times 2=144$. Hence $x y z$ equals either 12 or -12 .
Because $12-3 \sqrt{7}0$. Hence $x y z>0$. We conclude that $x y z=12$.