Let $$x = \sqrt[6]{27} - \sqrt{6\frac{3}{4}}$$ and $$y = \frac{\sqrt{45} + \sqrt{605} + \sqrt{245}}{\sqrt{80} + \sqrt{125}}$$, then the value of $$x^2 + y^2$$ is:

$$x = \sqrt[6]{27} - \sqrt{6\frac{3}{4}}$$

$$x = \sqrt{3} - \sqrt{\frac{27}{4}}$$

$$x = \sqrt{3} - \frac{3}{2} \sqrt{3}$$

$$x = - \frac{\sqrt{3}}{2}$$

$$x^2 = 3/4$$

$$y = \frac{\sqrt{45} + \sqrt{605} + \sqrt{245}}{\sqrt{80} + \sqrt{125}}$$

$$y = \frac{\sqrt{5 \times 9} + \sqrt{121 \times 5} + \sqrt{49 \times 5}}{\sqrt{16 \times 5} + \sqrt{5 \times25}}$$
$$y = \frac{3\sqrt{5} + 11\sqrt{5} + 7\sqrt{5}}{4\sqrt{ 5} + 5\sqrt{5}}$$

$$y = \frac{21\sqrt{5}}{9\sqrt{ 5}}$$

$$y^2 =\frac{2205}{405} = \frac{49}{9}$$

$$x^2 + y^2 = \frac{3}{4} + \frac{49}{9} = \frac{196 + 27}{36} = \frac{223}{36}$$