2019 SMC Problem 24

$\begin{aligned} x & = \sqrt{12-3\sqrt7} - \sqrt{12+3\sqrt 7} & \small \blue{\text{Note that }x < 0} \\ & = - \left(\sqrt{12+3\sqrt7} - \sqrt{12-3\sqrt 7} \right) \\ & = - \sqrt{\left(\sqrt{12+3\sqrt 7} - \sqrt{12-3\sqrt 7}\right)^2} \\ & = - \sqrt{12+3\sqrt 7 - 2\sqrt{12^2 - (3\sqrt 7)^2} + 12 - 3\sqrt 7} \\ & = - \sqrt{24-2 \times 9} = - \sqrt 6 \end{aligned}$

Similarly, $y = \sqrt{7-4\sqrt 3} - \sqrt{7+4\sqrt 3} = - \sqrt{2 \times 7-2\sqrt{49-16 \times 3}} = - \sqrt{12}$ and $z = \sqrt{2+\sqrt 3} - \sqrt{2-\sqrt 3} = \sqrt{2\times 2-2\sqrt{4-3}} = \sqrt 2$ .

Therefore $xyz = \sqrt{-6 (-12)(2)} = \boxed{12}$ .

$x$ , $y$ and $z$ are of the form $\sqrt{a-b} - \sqrt{a+b}$ for some $a$ and $b$ . Squaring, we get $(a-b)+(a+b) - 2\sqrt{(a-b)(a+b)} = 2a + \sqrt{a^2-b^2}$ So $\begin{aligned} x^2 &= 6 \\ y^2 &= 12 \\ z^2 &= 2 \\ \implies x^2y^2z^2 &= 144\end{aligned}$

From the initial expression we have $\begin{aligned} x&<0 \\ y&<0 \\ z&>0 \\ \implies xyz &> 0 \\ \therefore xyz &= 12 \end{aligned}$