Cubed Square Roots

Solution 1: Factoring the sum of cubes, we have

$\begin{aligned} (2 + \sqrt{3})^3 + (2 - \sqrt{3})^3 &= \left((2 + \sqrt{3}) + (2 - \sqrt{3})\right)\left((2 + \sqrt{3})^2 - (2 + \sqrt{3})(2 - \sqrt{3}) + (2 - \sqrt{3})^2\right) \\ &= (4)\left((4 + 4\sqrt{3} + 3) -(4 - 3) + (4 - 4\sqrt{3} + 3) \right) \\ &= (4)(13) \\ &= 52 \\ \end{aligned}$

Solution 2: $\begin{aligned} (2+\sqrt{3} ) ^3 &= ( 2 + \sqrt{3})^2 \times ( 2 + \sqrt{3} ) \\ &= ( 4 + 2 \cdot 2 \cdot \sqrt{3} + 3) \times ( 2 + \sqrt{3} ) \\ &= ( 7 + 4 \sqrt{3} ) (2 + \sqrt{3} ) \\ &= 14 + 8 \sqrt{3} + 7 \sqrt{3} +12 \\ &= 26 + 15 \sqrt{3} \\ \end{aligned}$

Similarly,

$\begin{aligned} (2-\sqrt{3} ) ^3 &= ( 2 - \sqrt{3})^2 \times ( 2 - \sqrt{3} ) \\ &= ( 4 - 2 \cdot 2 \cdot \sqrt{3} + 3) \times ( 2 - \sqrt{3} ) \\ &= ( 7 - 4 \sqrt{3} ) (2 - \sqrt{3} ) \\ &= 14 - 8 \sqrt{3} - 7 \sqrt{3} +12 \\ &= 26 - 15 \sqrt{3} \\ \end{aligned}$

Hence, $( 2 + \sqrt{3})^3 + ( 2 - \sqrt{3})^3 = (26 + 15 \sqrt{3} ) + (26 - 15 \sqrt{3}) = 52$ .