Find the value of
$$\frac{\sin^{6}15^{\circ} + \sin^{6}75^{\circ} + 6\sin^{2}15^{\circ}\sin^{2}75^{\circ}}{\sin^{4}15^{\circ} + \sin^{4}75^{\circ} + 5\sin^{2}15^{\circ}\sin^{2}75^{\circ}}$$

$$Let\ \sin^215^{\circ}\ =a\ and\ \sin^275^{\circ}\ =\ b\ $$

Then the given equation becomes,

$$\ \frac{\ a^3+b^3+6ab}{a^2+b^2+5ab}$$ 

=$$\ \frac{\ \left(a+b\right)^3-3ab\left(a+b\right)+6ab}{\left(a+b\right)^2-2ab+5ab}$$     $$\ \therefore\ \ a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)$$ , $$\ \therefore\ \ a^2+b^2=\left(a+b\right)^2-2ab$$

Using $$\sin\left(\theta\right)\ =\cos\left(90-\theta\ \ \right)$$

we get, $$\sin^275^{\circ\ }=\cos^215^{\circ\ }$$

now, a+b = $$\sin^215^{\circ\ }+\sin^275^{\circ\ }$$ = $$\sin^215^{\circ\ }+\cos^215^{\circ\ }$$ = 1.

The equation becomes, $$=\ \frac{\ 1-3ab+6ab}{1-2ab+5ab}\ =\ \frac{\ 1+3ab}{1+3ab}\ =1\ $$