1(1−3)2\;\frac{1}{(1-\sqrt{3})^{2}}(1−3)21
(1−3)2=(1−3)(1−3)(1-\sqrt{3})^{2} = (1-\sqrt{3})(1-\sqrt{3})(1−3)2=(1−3)(1−3)
1−23+3=4−231 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}1−23+3=4−23
14−23\;\frac{1}{4-2\sqrt{3}}4−231
After rationalising (multiplying the denominator and numerator with 4+234+2\sqrt{3}4+23, we have
4+234=1+ 123\;\frac{4+2\sqrt{3}}{4} = 1 + \;\frac{1}{2}\sqrt{3}44+23=1+213
