Question

Solve: $\frac{3+\sqrt{3}}{3-\sqrt{3}}+\frac{3-\sqrt{3}}{3+\sqrt{3}}+\frac{1}{3+\sqrt{3}}-\frac{1}{3-\sqrt{3}}$

Answer 1

We have,

$(\frac{3+\sqrt{3}}{3-\sqrt{3}}+\frac{3-\sqrt{3}}{3+\sqrt{3}})+(\frac{1}{3+\sqrt{3}}-\frac{1}{3-\sqrt{3}})$

[Using cross multiplication method, $\frac{a}{b}+\frac{c}{d}=\frac{(a\times d)+(b\times c)}{(b\times d)}$]

$=\frac{(3+\sqrt{3})(3+\sqrt{3})+(3-\sqrt{3})(3-\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}+\frac{1}{3+\sqrt{3}}-\frac{1}{3-\sqrt{3}}$

$=\frac{(3+\sqrt{3}{)}^2+(3-\sqrt{3}{)}^2}{(3-\sqrt{3})(3+\sqrt{3})}+\frac{(3-\sqrt{3})-(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}$

$=\frac{{(3+\sqrt{3})}^2+{(3-\sqrt{3})}^2}{{\left(3\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{3-\sqrt{3}-3-\sqrt{3}}{{\left(3\right)}^2-{\left(\sqrt{3}\right)}^2}$

[Using the identity, $(a+b{)}^2=(a^2+b^2+2ab),(a-b{)}^2=(a^2+b^2-2ab),$

and $(a^2-b^2)=(a-b)(a+b)$ ]

$=\frac{(3^2+3+6\sqrt{3})+(3^2+3-6\sqrt{3})}{9-3}+\frac{-2\sqrt{3}}{9-3}$

$=\frac{(12+6\sqrt{3})+(12-6\sqrt{3})}{(9-3)}+\frac{(-2\sqrt{3})}{(9-3)}$

$=\frac{24}{6}-\frac{2\sqrt{3}}{6}$

$=4-\frac{\sqrt{3}}{3}$

$=\frac{12-\sqrt{3}}{3}$