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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.
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