The only way I can show
$$\cos(20^\circ)\cos(30^\circ)\cos(40^\circ)=\cos^2(10^\circ)\cos(50^\circ)$$
stinks of sweat and brute force:
If $\cos(10^\circ)=x$ then using $\cos(a+b) = \cos a \cos b – \sin a \sin b$ and $\cos^2 + \sin^2 = 1$ you get
$$
\cos(20^\circ) = 2x^2-1\\
\cos(30^\circ) = 4x^3-3x \\
\cos(40^\circ) = 8x^4-8x^2+1\\
\cos(80^\circ) = 128x^8-256x^6+160x^4-32x^2+1
$$
Then
$$
\sin(10^\circ) = 128x^8-256x^6+160x^4-32x^2+1
$$
and using $\sin (a+b) = \sin(a) \cos(b) + \cos a \sin b$ and $\sin(30^\circ)=\frac12$ you get
$$
\sin(20^\circ)= 256x^9 -512x^7+320x^5-64x^3+2x \\
\cos(50^\circ)=(2x^2-1)(4x^x-3x) – \frac12(256x^9 -512x^7+320x^5-64x^3+2x)\\
\cos(50^\circ) = -128x^9+256x^7-152x^5+22x^3+2x
$$
Now that we have all the needed values of cosines, we have to prove that
$$
(2x^2-1)(4x^x-3x)(8x^4-8x^2+1)=x^2(-128x^9+256x^7-152x^5+22x^3+2x)\\
64x^9-144x^7+112x^5-34x^3+3x=-128x^{11}+256x^9-152x^7+22x^5+2x^3
$$
The tool we have is that $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ so
$$x^3=\frac{\sqrt{3}/2+3x}{4}
$$
We now have to use that replace the highest power of $x$ by 2 powers lower; for example, using
$$x^{11} = \frac{3x^9}{4} + \frac{\sqrt{3}}{8}x^8$$
I had to do this replacement trick thirteen times (seven on the right side and six on the left) before I came to
$$
-x^3 + \frac{\sqrt{3}x^2}{2} + \frac{3x}{4} = -x^3 + \frac{\sqrt{3}x^2}{2} + \frac{3x}{4}
$$
And we come to the original question — there has got to be an easier way to show this!
Best Answer
Using degrees as a unit, $$ \cos20^{\circ}\cos30^{\circ} = \frac{1}{2}\left(\cos50^{\circ}+\cos10^{\circ}\right)$$ and by multiplying both sides by $\cos40^{\circ}$: $$ \cos20^{\circ}\cos30^{\circ}\cos40^{\circ} = \frac{1}{4}\left(\cos90^{\circ}+\cos10^{\circ}+\cos50^{\circ}+\cos30^{\circ}\right). $$ while the same approach leads to: $$ \cos10^{\circ}\cos10^{\circ}\cos50^{\circ} = \frac{1}{4}\left(\cos70^{\circ}+2\cos50^{\circ}+\cos30^{\circ}\right)$$ so it is enough to prove that: $$ \cos70^{\circ}+\cos50^{\circ}=\cos10^{\circ} $$ that follows from $\cos60^{\circ}=\frac{1}{2}$.