Find the value of $\alpha,\beta$ for the equation $\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}$ Find the value of $\alpha,\beta$ for the equation $\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}$ 
$\alpha>0$ & $\beta<\frac{\pi}{2}$
I get the following step after some substitution
$\cos2\alpha + \cos2\beta+\cos2(\alpha+\beta)=-\frac{3}{2}$ from here not able to proceed.

Suppose $$0<\alpha,\beta<\frac\pi2.\tag{1}$$ From 
$$\cos2\alpha + \cos2\beta+\cos2(\alpha+\beta)=-\frac{3}{2}$$
one has
$$ \cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)=\frac34. $$
Note that
$$\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}\tag{2}$$
implies
$$\cos^2\alpha \cos^2\beta \cos^2(\alpha +\beta)=\frac{1}{64}.$$
By the AM-GM inequality
$$ a+b+c\ge3\sqrt[3]{abc}$$
one has
$$ \frac34=\cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)=3\sqrt[3]{\cos^2\alpha\cos^2\beta\cos^2(\alpha+\beta)}=3\sqrt[3]{\frac{1}{64}}=\frac34 $$
and the equal sign holds if and only if
$$\cos^2\alpha=\cos^2\beta=\cos^2(\alpha+\beta).\tag{3}$$
From (1)(2)(3), it is easy to see
$$ \alpha=\beta=\frac{\pi}{3}. $$
