I will give you some hints toward the solution;
I don't know what the exact translation is, but i hope you know what I mean if I say I write C p q as the combination of 'p above q'
Then the binomial expansion is given as (a+b) n = C n na nb 0 + C n (n-1)a (n-1)b+C n (n-2)a (n-2)b 2+....+C n 0a 0b n
For each term we can write C n ra (n-r)b r (basically, the above is the sum of this general term for r is all integers between 0 and n)
So what we want to do is make C n ra (n-r)b r constant for (x-1/x) 4
We take a = x, b = -1/x and n = 4. Then we need to find an r which gives an x 0 in C n ra (n-r)b r so that this general term no longer depends on x.
so try filling in these variables in C n ra (n-r)b r and rewrite it such that it looks somewhat like this x something with r * some other function without x and then equate 'something with r' = 0 to find the value of r. Then fill in r in the general term and you have the constant.
Hope that will work for you.
+2353 
