3 WITHOUT using a calculator, show that \( \frac{9-\sqrt{54}}{6 \sqrt{2}} \) is equal to \( \frac{3 \sqrt{2}-2 \sqrt{3}}{4} \).

To show that \( \frac{9-\sqrt{54}}{6 \sqrt{2}} = \frac{3 \sqrt{2}-2 \sqrt{3}}{4} \), we will simplify the left-hand side. Starting with \( \sqrt{54} \): \[ \sqrt{54} = \sqrt{9 \cdot 6} = \sqrt{9} \cdot \sqrt{6} = 3\sqrt{6} \] Substituting this back into the left-hand side gives: \[ \frac{9 - \sqrt{54}}{6 \sqrt{2}} = \frac{9 - 3\sqrt{6}}{6 \sqrt{2}} \] Now we can split the fraction: \[ = \frac{9}{6 \sqrt{2}} - \frac{3\sqrt{6}}{6 \sqrt{2}} = \frac{3}{2\sqrt{2}} - \frac{\sqrt{6}}{2\sqrt{2}} \] Next, let's simplify \(\frac{3}{2\sqrt{2}}\) and \(\frac{\sqrt{6}}{2\sqrt{2}}\): \[ \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4} \] \[ \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{6}}{\sqrt{2} \cdot 2} = \frac{\sqrt{6}}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} \cdot \sqrt{2}}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \] Thus, combining these results: \[ \frac{3}{2\sqrt{2}} - \frac{\sqrt{6}}{2\sqrt{2}} = \frac{3\sqrt{2}}{4} - \frac{\sqrt{3}}{2} \] Thankfully, we can express \(\frac{\sqrt{3}}{2}\) with a common denominator: \[ = \frac{3\sqrt{2}}{4} - \frac{2\sqrt{3}}{4} = \frac{3\sqrt{2} - 2\sqrt{3}}{4} \] Finally, we see that the two sides are equal: \[ \frac{9-\sqrt{54}}{6 \sqrt{2}} = \frac{3\sqrt{2}-2\sqrt{3}}{4} \] Therefore, we have shown that \( \frac{9-\sqrt{54}}{6 \sqrt{2}} \) is indeed equal to \( \frac{3 \sqrt{2}-2 \sqrt{3}}{4} \).