Question: Rationalise the denominator of
$\frac{(6 + \sqrt{3})(6 - \sqrt{3})}{\sqrt{33}}$
1. **State the problem:** Rationalise the denominator of the expression $$\frac{(6 + \sqrt{3})(6 - \sqrt{3})}{\sqrt{33}}$$.
2. **Recall the formula:** To rationalise a denominator containing a square root, multiply numerator and denominator by the conjugate or the radical itself to eliminate the root in the denominator.
3. **Simplify the numerator first:** Use the difference of squares formula:
$$ (a+b)(a-b) = a^2 - b^2 $$
Here, $$a=6$$ and $$b=\sqrt{3}$$, so
$$ (6 + \sqrt{3})(6 - \sqrt{3}) = 6^2 - (\sqrt{3})^2 = 36 - 3 = 33 $$
4. **Rewrite the expression:**
$$ \frac{33}{\sqrt{33}} $$
5. **Rationalise the denominator:** Multiply numerator and denominator by $$\sqrt{33}$$:
$$ \frac{33}{\sqrt{33}} \times \frac{\sqrt{33}}{\sqrt{33}} = \frac{33 \times \sqrt{33}}{\sqrt{33} \times \sqrt{33}} $$
6. **Simplify the denominator:**
$$ \sqrt{33} \times \sqrt{33} = 33 $$
7. **Substitute back:**
$$ \frac{33 \times \sqrt{33}}{33} $$
8. **Cancel common factors:**
$$ \frac{\cancel{33} \times \sqrt{33}}{\cancel{33}} = \sqrt{33} $$
**Final answer:**
$$ \sqrt{33} $$