Rationalise the denonator \( \frac{6-4 \sqrt{3}}{6+4 \sqrt{3}} \)

Calculate the value by following steps: - step0: Calculate: \(\frac{\left(6-4\sqrt{3}\right)}{\left(6+4\sqrt{3}\right)}\) - step1: Remove the parentheses: \(\frac{6-4\sqrt{3}}{6+4\sqrt{3}}\) - step2: Rewrite the expression: \(\frac{2\left(3-2\sqrt{3}\right)}{2\left(3+2\sqrt{3}\right)}\) - step3: Reduce the fraction: \(\frac{3-2\sqrt{3}}{3+2\sqrt{3}}\) - step4: Multiply by the Conjugate: \(\frac{\left(3-2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\) - step5: Multiply the numbers: \(\frac{\left(3-2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}{-3}\) - step6: Multiply the numbers: \(\frac{21-12\sqrt{3}}{-3}\) - step7: Rewrite the expression: \(\frac{3\left(7-4\sqrt{3}\right)}{-3}\) - step8: Reduce the fraction: \(\frac{7-4\sqrt{3}}{-1}\) - step9: Calculate: \(4\sqrt{3}-7\) To rationalize the denominator of the expression \[ \frac{6-4 \sqrt{3}}{6+4 \sqrt{3}}, \] we multiply both the numerator and the denominator by the conjugate of the denominator, which is \(6 - 4 \sqrt{3}\). 1. **Multiply the numerator and denominator:** \[ \frac{(6-4 \sqrt{3})(6-4 \sqrt{3})}{(6+4 \sqrt{3})(6-4 \sqrt{3})}. \] 2. **Calculate the denominator:** Using the difference of squares formula, we have: \[ (6+4 \sqrt{3})(6-4 \sqrt{3}) = 6^2 - (4 \sqrt{3})^2 = 36 - 48 = -12. \] 3. **Calculate the numerator:** Now, we expand the numerator: \[ (6-4 \sqrt{3})(6-4 \sqrt{3}) = 6^2 - 2 \cdot 6 \cdot 4 \sqrt{3} + (4 \sqrt{3})^2 = 36 - 48 \sqrt{3} + 48 = 84 - 48 \sqrt{3}. \] 4. **Combine the results:** Now we have: \[ \frac{84 - 48 \sqrt{3}}{-12}. \] 5. **Simplify the expression:** Dividing each term in the numerator by \(-12\): \[ \frac{84}{-12} - \frac{48 \sqrt{3}}{-12} = -7 + 4 \sqrt{3}. \] Thus, the rationalized form of the expression is: \[ 4 \sqrt{3} - 7. \]