Evaluate: $$\left[\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right]^2  + \left[\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\right]^2$$

Using the identities

$$ (a + b)^2 = a^2 + 2ab + b^2 $$

$$ (a - b)^2 = a^2 - 2ab + b^2 $$

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

Rationalizing the denominator,

$$\left[\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right] = \frac{\sqrt{3} + 1}{\sqrt{3} -1} \times \frac{\sqrt{3} + 1}{\sqrt{3} +1} $$

Solving the equation using identities we get

$$ \frac{\sqrt{3} + 1}{\sqrt{3} -1} \times \frac{\sqrt{3} + 1}{\sqrt{3} +1} = \frac{4 + 2\sqrt3}{2} $$

            = $$ 2 + \sqrt 3 $$

$$\left[\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right]^2 =  (2 + \sqrt 3)^2$$

= $$ 7 + 4\sqrt3 $$

Rationalizing the denominator,

$$\left[\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\right] = \frac{\sqrt{3} - 1}{\sqrt{3} +1} \times \frac{\sqrt{3} -1}{\sqrt{3} -1} $$

Solving the equation using identities we get

$$ \frac{\sqrt{3} -1}{\sqrt{3} +1} \times \frac{\sqrt{3} - 1}{\sqrt{3} -1} = \frac{4 - 2\sqrt3}{2} $$

            = $$ 2 - \sqrt 3 $$

$$\left[\frac{\sqrt{3} -1}{\sqrt{3} + 1}\right]^2 =  (2 - \sqrt 3)^2$$

= $$ 7 - 4\sqrt3 $$

Thus,

$$\left[\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right]^2 + \left[\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\right]^2 = 7 + 4\sqrt 3 + 7 - 4\sqrt 3 $$

   = 14