If $$A = 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta)$$ then the value of $$3 \alpha$$ such that $$\cos \alpha = \sqrt{\frac{3 + A}{5 + A}}$$ is:

Given,  $$A=2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$$

$$=$$>  $$A=2\left(\left(\sin^2\theta\right)^3+\left(\cos^2\theta\right)^3\right)-3(\sin^4\theta+\cos^4\theta)$$

$$=$$>  $$A=2\left(\sin^2\theta+\cos^2\theta\right)\left(\sin^4\theta\ -\sin^2\theta\ \cos^2\theta\ +\cos^4\theta\ \right)-3(\sin^4\theta+\cos^4\theta)$$

$$=$$>  $$A=2\left(\sin^4\theta\ -\sin^2\theta\ \cos^2\theta\ +\cos^4\theta\ \right)-3(\sin^4\theta+\cos^4\theta)$$

$$=$$>  $$A=2\sin^4\theta\ -2\sin^2\theta\ \cos^2\theta\ +2\cos^4\theta\ -3\sin^4\theta-3\cos^4\theta$$

$$=$$>  $$A=-\sin^4\theta-\cos^4\theta-2\sin^2\theta\ \cos^2\theta\ $$

$$=$$>  $$A=-\left(\sin^4\theta+\cos^4\theta+2\sin^2\theta\ \cos^2\theta\ \right)$$

$$=$$>  $$A=-\left(\sin^2\theta+\cos^2\theta\ \right)^2$$

$$=$$>   $$A=-1$$

$$\therefore\ $$ $$\cos\alpha=\sqrt{\frac{3+A}{5+A}}=\sqrt{\frac{3-1}{5-1}}$$

$$=$$>  $$\cos\alpha=\sqrt{\frac{2}{4}}$$

$$=$$>  $$\cos\alpha=\sqrt{\frac{1}{2}}$$

$$=$$>  $$\cos\alpha=\frac{1}{\sqrt{2}}$$

$$=$$>  $$\cos\alpha=\cos45^{\circ\ }$$

$$=$$>  $$\alpha=45^{\circ\ }$$

$$=$$>  $$3\alpha=3\times45^{\circ\ }=135^{\circ\ }$$

Hence, the correct answer is Option D