Question: In Dirac's Principles of QM following is stated:

$$
\langle x | A + B | x \rangle = \langle x | A | x \rangle + \langle x | B |x \rangle
$$

but

$$
\langle x | AB | x \rangle \ne \langle x | A | x \rangle \langle x | B | x \rangle
$$

and so $\langle x|A|x \rangle$ is not exact but average value of observable A otherwise in second relation both sides would have to be equal.
I don't understand the second relation. Shouldn't both side be equal like this,
$\langle x|AB|x \rangle = \langle x|A(B|x \rangle) = b \langle x|A|x \rangle = ba \langle x|x \rangle = ba = \langle x|A|x \rangle \langle x|B|x \rangle$ , where $a$ and $b$ are corresponding eigenvalues. What is wrong here?

Edit: It is embarrassing. I indeed was thinking $|x\rangle$ was eigenvector of both $A$ and $B$ out of sleep deprivation I suppose which I only realised this morning. So anyway I am going to ask moderator to delete this question.

Answer: 
The coin is denser than water. When placed on the box, it will displace it's weight in water, which is more volume than the coin. When placed directly in the water it will sink to the bottom and displace only it's own volume in water, which will be less because it's denser.

Another way to think about it is to have a platform below the box open to the water and a coin on the bottom of the tank. If you place the coin on the platform, it will pull the box down and displace more water than when it was on the bottom of the tank. It will actually displace the same amount of water as if you put it on top of the box.

[![enter image description here](https://i.stack.imgur.com/iWMY8.png)](https://i.stack.imgur.com/iWMY8.png)
I'm not an expert so please check if this makes sense.

It seems like the "reason" for this is that when it's on the box, it's displacing water "using" the box. Since the box is less dense, the same weight occupies more space, so the weight of the coin gets amplified by the ratio of its specific weight to the box's specific weight. For example, if the box's specific weight is 10 times lighter than the coin's, the coin will now be pushing on 10 times the volume as its own (- the coin's) volume into the water. (Ignoring the difference in water level. I'm just making the point that what is pushed is "box", not "coin".)