If $$\sin \theta + \cos \theta = m$$, then $$\sin^{6} \theta + \cos^{6}\theta$$ equals

$$\sin \theta + \cos \theta = m$$

Squaring both sides, we get

$$\sin^2\theta\ +\cos^2\theta\ +2\sin\theta\ \cos\theta\ \ =\ m^2$$

We know that $$\sin^2\theta\ +\cos^2\theta\ =1$$

$$1+2\sin\theta\ \cos\theta\ \ =\ m^2$$

$$\sin\theta\ \cos\theta\ \ =\ \dfrac{m^2-1}{2}$$  -----(1)

Now, using the algebra property $$a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)$$

Assuming $$a=\sin^2\theta\ $$ and $$b=\cos^2\theta\ $$

$$\sin^6\theta\ +\cos^6\theta\ =\left(\sin^2\theta\ +\cos^2\theta\ \right)^3-3\sin^2\theta\ \cos^2\theta\ \left(\sin^2\theta\ +\cos^2\theta\ \right)$$

$$\sin^6\theta\ +\cos^6\theta\ =\ 1-3\sin^2\theta\ \cos^2\theta\ \left(1\ \right)$$

Substituting the value from equation (1)

$$\sin^6\theta\ +\cos^6\theta\ =\ 1-3\left(\dfrac{m^2-1}{2}\right)^2$$

Thus, option D is the correct answer.