A 3-input LUT is an 8-bit memory, and that memory can contain one of 28 = 256 different values. Each value represents one possible Boolean function of the three input variables.
A 2-input LUT can contain one of 24 = 16 different functions.
When you cascade two 2-input LUTs, you are effectively creating a temporary variable T that is a function of two of the inputs (16 choices) and then the final output is a function of T and the third input (16 more choices), for a total of 32 choices. Since you get to pick which of the three input variables is connected to the second LUT, you can actually create up to 32 × 3 = 96 different functions this way.
However, this leaves 256 - 96 = 160 functions that cannot be created. These additional functions require a true 3-input LUT, or two 2-input LUTs plus a mux.
It's easy to construct such a function — a function that requires two different 2-input LUTs. For example, suppose you want A AND B when C is high, but A OR B when C is low:
$$Y = ABC + (A+B)\overline{C} = ABC + A\overline{C}+B\overline{C}$$
There's no way to decompose this into two functions of two variables each. If you look at the 3-D Karnaugh map of this function, you'll gain some insight as to why.
If all you have is 2-input LUTs, you'll need to use three of them to create a 2:1 mux, and then two more to hold your 8 function bits, for a total of five 2-input LUTs to emulate a general-purpose 3-input LUT.
If you are given
$$Y = f(A, B, C)$$
The five 2-input LUTS are
$$S = f(A, B, 0)$$
$$T = f(A, B, 1)$$
$$U = S \cdot \overline{C}$$
$$V = T \cdot C$$
$$Y = U + V$$
Here's the same thing in the form of a diagram:
simulate this circuit – Schematic created using CircuitLab
