If $$10\sin^4\theta + 15\cos^4\theta = 6$$, then the value of $$\dfrac{27\csc^6\theta + 8\sec^6\theta}{16\sec^8\theta}$$ is:

Let $$s = \sin^2\theta$$ and $$c = \cos^2\theta$$. Then $$s + c = 1$$ since $$\sin^2\theta + \cos^2\theta = 1$$.

The given equation is $$10\sin^4\theta + 15\cos^4\theta = 6$$. Re-write it using $$s$$ and $$c$$:
$$10s^2 + 15c^2 = 6$$.

Because $$c = 1 - s$$, substitute to get a quadratic in $$s$$:
$$10s^2 + 15(1 - s)^2 = 6$$

Simplify:
$$10s^2 + 15(1 - 2s + s^2) = 6$$
$$10s^2 + 15 - 30s + 15s^2 = 6$$
$$25s^2 - 30s + 15 - 6 = 0$$
$$25s^2 - 30s + 9 = 0$$ $$-(1)$$

Compute the discriminant of $$(1)$$:
$$\Delta = (-30)^2 - 4 \times 25 \times 9 = 900 - 900 = 0$$.

Since $$\Delta = 0$$, there is one repeated root:
$$s = \frac{30}{2 \times 25} = \frac{3}{5}$$.

Therefore
$$\sin^2\theta = s = \frac{3}{5}\quad\text{and}\quad \cos^2\theta = 1 - s = \frac{2}{5}$$.

Compute the higher-power reciprocals needed.
$$\csc^6\theta = \frac{1}{\sin^6\theta} = \frac{1}{\left(\dfrac{3}{5}\right)^3} = \frac{1}{\dfrac{27}{125}} = \frac{125}{27}$$
$$\sec^6\theta = \frac{1}{\cos^6\theta} = \frac{1}{\left(\dfrac{2}{5}\right)^3} = \frac{1}{\dfrac{8}{125}} = \frac{125}{8}$$
$$\sec^8\theta = \frac{1}{\cos^8\theta} = \frac{1}{\left(\dfrac{2}{5}\right)^4} = \frac{1}{\dfrac{16}{625}} = \frac{625}{16}$$

Form the required numerator:
$$27\csc^6\theta + 8\sec^6\theta = 27\left(\frac{125}{27}\right) + 8\left(\frac{125}{8}\right) = 125 + 125 = 250$$

Form the required denominator:
$$16\sec^8\theta = 16\left(\frac{625}{16}\right) = 625$$

Hence
$$\dfrac{27\csc^6\theta + 8\sec^6\theta}{16\sec^8\theta} = \dfrac{250}{625} = \dfrac{2}{5}$$.

The value equals $$\dfrac{2}{5}$$, which matches Option A.