1. Interpret the notation

The expression means $15\sin^4\alpha + 10\cos^4\alpha = 6.$ We need to find $27\sec^6\alpha + 8\csc^6\alpha.$

2. Let $x = \sin^2\alpha, \qquad y = \cos^2\alpha.$ Then $x+y=1,$ and $\sin^4\alpha = x^2, \qquad \cos^4\alpha = y^2.$ So the given equation becomes $15x^2 + 10y^2 = 6.$ Using $y=1-x$, $15x^2 + 10(1-x)^2 = 6.$

3. Solve for $x$

Expand: $15x^2 + 10(1-2x+x^2)=6$ $15x^2 + 10 - 20x + 10x^2 = 6$ $25x^2 - 20x + 4 = 0.$

Now factor: $25x^2 - 20x + 4 = (5x-2)^2 = 0.$ Thus, $x = \frac{2}{5}.$ Hence, $\sin^2\alpha = \frac{2}{5}, \qquad \cos^2\alpha = 1-\frac{2}{5}=\frac{3}{5}.$

4. Find $\sec^6\alpha$ and $\csc^6\alpha$

Since $\cos^2\alpha = \frac{3}{5},$ we get $\sec^2\alpha = \frac{5}{3} \implies \sec^6\alpha = \left(\frac{5}{3}\right)^3 = \frac{125}{27}.$

Also, $\sin^2\alpha = \frac{2}{5},$ so $\csc^2\alpha = \frac{5}{2} \implies \csc^6\alpha = \left(\frac{5}{2}\right)^3 = \frac{125}{8}.$

5. Compute the required value