How to calculate the GCD? How to evaluate the following with the help of Mobius function ?
$$\displaystyle\sum_{i=1}^n  \sum_{j=i+1}^n     \sum_{k=j+1}^n   \sum_{l=k+1}^n  {gcd(i,j,k,l)^4} .$$
In other words, we have to select all possible quadruplets from (1 ton n), and then sum up their value with power 4.
Example:-
N=4 :
(1,2,3,4) :   gcd(1,2,3,4)^4 = 1
Total sum = 1
For Second  Case, let N=5 :
(1,2,3,4) :   gcd(1,2,3,4)^4 = 1
(1,2,3,5) :   gcd(1,2,3,5)^4 = 1
(1,2,4,5) :   gcd(1,2,4,5)^4 = 1
(1,3,4,5) :   gcd(1,3,4,5)^4 = 1
(2,3,4,5) :   gcd(2,3,4,5)^4 = 1
Total sum = 1+1+1+1+1 = 5.
My approach is:
To calculate the number of quadruplets with gcd=2,3,4,...till n.
Say, the number of quadruplets with gcd=2 are x1, for gcd=3, its x2...and so on.
Now,  l  = $$\binom{n}{4}$$-(x1+x2+.......xn-1) = number of quadruplets with gcd=1. 
Now, the final answer = l^4+x1^4+x2^4+.....
The only problem I have is how to calculate the number of quadruplets with gcd=x with the help of mobius function ?

A quadruplet with gcd a multiple of $d$ occurs iff all $i,j,k,l$ are multiples of $d$, for which there are ${\lfloor n/d\rfloor\choose d}$ possibilities.
So you'll arrive at
$${n\choose 4}\cdot 1^4+{\lfloor n/2\rfloor\choose 4}\cdot (2^4-1^4) +{\lfloor n/3\rfloor\choose 4}\cdot (3^4-1^4)+{\lfloor n/4\rfloor\choose 4}\cdot (4^4-2^4)+\cdots$$
