Confusion regarding calculating shortest path in a cube An ant is at a corner of a cubical room of side a. The ant can move with constant speed u. The minimum time taken to reach the farthest corner of the cube is
I am completely clueless how is it possible to determine the shortest path which the insert can as insects can't fly in this case. It has to walk on the cube.

This is the book's solution 
I will be grateful if you help me
Thanks
 A: The shortest distance between any 2 points is given by the line joining the 2 points.
If the ant could fly, the shortest distance would have been given by the body diagonal of the cube, which is $\bf \sqrt3a$.
But ants generally don't fly (thank god!), ants walk. So, the shortest distance between 2 points in a cube is hard to see as such. 
But if we unfold the cube (as suggested in the comments) then all the points on the cube will now lie on a plane and it's easy to see then that again the shortest path between the 2 points is given by the straight line joining them.

As you can see in the pic above, that the shortest distance between $PR$ is given by the line joining them, and using Pythagoras theorem, we find it's length as follows:
$$PR^2=(2a)^2 + a^2$$
$$\Rightarrow \boxed {PR=\sqrt5a}$$

Even if it were a cuboid, you would have proceeded in a similar manner. Just be careful as there you'll have different combinations as you travel along whether $(lb)$ or $(bh)$ or $(hl)$. Perhaps you'll also notice which combination would always give the shortest route.
Also, note that the above image is not the only way to unfold the cube, it could be done in any of the following ways (and obviously, they would all give the same answer, use whichever you feel easy to work with):


I have addressed the main thing, but to get the final answer, you are asked to find the time taken and $\text{Speed}=\text{Distance}/\text{Time}$.
So, time taken, $\bf t$ $=\frac ds=$$\bf\frac {\sqrt5a}u$.
