8=56 7=42 6=30 5=20 3=?

It turns out that the '?' can be whatever we want it to be!

Allow me to illustrate. Take for example, f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42, and evaluate f(8), f(7), f(6), f(5), f(3). It turns out that:

f(8)=56,
f(7)=42, 
f(6)=30, 
f(5)=20, 
f(3)=9.

Wait, what sorcery is this? Turns out that although the popular rule f(x)=x(x-1), which gives f(x)=6, satisfies the known values in the sequence, that f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42 also satisfies them--except with a different value of f(3)!

Here's another one that also works but gives f(3)=12:
f(x)=(1/20)x^4-(13/10)x^3+(271/20)x^2-(543/10)x+84

And here is one where f(3)=π
f(x)=(1/120)(π-6)x^4-(13/60)(π-6)x^3+(1/120)(251π-1386)x^2+(1/60)(3138-533π)x+14(π-6)

Finally, in general if you want the '?'=k, i.e., f(3)=k where k is the value of your choice, then
(1/120)(k-6)x^4-(13/60)(k-6)x^3+(1/120)(251k-1386)x^2+(1/60)(3138-533k)x+14(k-6)

More details here:
https://www.scribd.com/doc/260182194/Elementary-Sequences

The answer is 6.

Possibly as follows:

7x8 = 56

6x7 = 42

5x6 = 30

4x5 = 20

3x4 = 12

.