Conratulations to Sue Liu, Jonathan and Tom of Madras College, St Andrew's and to Sanjay of The Perse School, Cambridge for their solutions to this problem. Here is Sanjay's solution.

$$\sqrt{8 -4\sqrt{3}} = \sqrt{a} - \sqrt{b}$$ The tactic I shall employ here will be to square both sides and solve for $a$ and $b$. $$8 -4\sqrt{3} = a - 2\sqrt{ab} + b$$. From this it is clear that the following equations must hold

\begin{eqnarray} a+b &=& 8 \\ 4\sqrt{3}
&=& 2 \sqrt{ab} \end{eqnarray}

These simultaneous equations can be solved for $a$ and $b$. From
equation 2. $$2 \sqrt{3} = \sqrt{ab}$$ From equation 1. $$b = 8
-a$$ Substituting equation 4 into 3 gives
\begin{eqnarray}\\ 2 \sqrt{3} &=&
\sqrt{a(8-a)}\\ 12 &=& a(8-a) \\ a^2 - 8a + 12 &=&
0 \\ (a-6)(a-2) &=& 0 \\ a &=& 2\ or \ a = 6
\end{eqnarray}

From the original equation it is clear that $a = 6$ and $b=2$
because the left-hand side must be positive. Hence $$
\sqrt{8-4\sqrt{3}} = \sqrt{6} - \sqrt{2}$$ This can easily be
generalised, as follows. $$\sqrt{x-y\sqrt{z}} = \sqrt{a} -
\sqrt{b}$$ Upon squaring both sides this yields the equations
\begin{eqnarray} a + b &=& x \\ ab
&=& \frac{y^2z}{4} \end{eqnarray}

From equation 5, it is clear that $$b = x - a$$ Substituting this
into equation 6 gives \begin{eqnarray} a(x-a) = \frac{y^2z}{4} \\
a^2 -ax + \frac{y^2z}{4} \end{eqnarray} Using the formula for
solving quadratics, and setting $a$ to be a larger than $b$ (as in
the original equation) gives the following solutions.
\begin{eqnarray} a &=& \frac{x+\sqrt{x^2
- y^2z}}{2} \\b &=& \frac{x-\sqrt{x^2 -
y^2z}}{2}\end{eqnarray}

Note that if the minuses in the folrmula are changed to pluses,
equations 5 and 6 will still be the same, and so you'll still get
the same answers.