If the angle between any two diagonals of a cube is $\theta$ then 

DC’s of diagonals OP and AQ are 
$\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}and-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{3}$ 
If angle between them is $\theta$ , then
$\cos \theta =\left(\frac{1}{\sqrt{3}}\right)(-\frac{1}{\sqrt{3}})+\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right)$ 
 
$=\frac{1}{3}$