Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$.

From the triple-angle formula $\sin (3\theta) = - 4\sin^3\theta + 3\sin\theta$ when $\sin (3\theta) = 1/2$, we get that $\sin(10^\circ)$, $\sin(50^\circ)$, $\sin(-70^\circ)$ are the roots of $8x^3-6x+1$. Therefore $$ \sin(10^\circ) \sin(50^\circ) \sin(70^\circ) =-\sin(10^\circ) \sin(50^\circ) \sin(-70^\circ) =-(-\frac18) =\frac18. $$

$$\begin{eqnarray*}\sin(A)\sin(B)\sin(C) &=& \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)\sin C\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)-\sin(C+A+B)-\sin(C-A-B)\right)\end{eqnarray*}$$ hence:

$$\begin{eqnarray*} \sin(10^\circ)\sin(50^\circ)\sin(70^\circ)&=&\frac{1}{4}\left(\sin(110^\circ)+\sin(30^\circ)+\sin(-10^\circ)-\sin(130^\circ)\right) \\&=&\frac{1}{4}\left(\cos(20^\circ)+\sin(30^\circ)-\sin(10^\circ)-\cos(40^\circ)\right)\end{eqnarray*}$$ but: $$ \cos(20^\circ)-\cos(40^\circ) = 2\sin(30^\circ)\sin(10^\circ) = \sin(10^\circ)$$ hence:

$$ \sin(10^\circ)\sin(50^\circ)\sin(70^\circ) = \color{red}{\frac{1}{8}}.$$

Here is a method that uses Gelfand's hint.

Let us first transform all the sines into cosines of the complementary angle:

$\tag{1}M:=\sin(10^\circ)\sin(50^\circ)\sin(70^\circ)=\cos(80^\circ)\cos(40^\circ)\cos(20^\circ)$

Let us multiply LHS and RHS of $(1)$ by $\sin(20^\circ):$

$M \sin(20^\circ)=\cos(80^\circ)\cos(40^\circ)(\cos(20^\circ)\sin(20^\circ).$

$=\cos(80^\circ)\cos(40^\circ)\frac12\sin(40^\circ).$

(applying the so useful formula $\sin(a)\cos(a)=\frac12\sin(2a).$)

Using the same trick again, we get:

$M \sin(20^\circ)=\frac14\cos(80^\circ)\sin(80^\circ).$

And once more:

$M \sin(20^\circ)=\frac18\sin(160^\circ).$

But $\sin(160^\circ)=\sin(20^\circ).$

We can thus conclude that

$M=\dfrac18.$

Edit: in fact, I just discover that it is a direct consequence of Morrie's law.