Dear Hrushikesh, This is an interesting question, If  we calculate [ x , p ] in terms of well known creation and annihilation operators in normal unit, i.e. [ x , p ] = i [ a , a+ ] and represent  a+ and a operators in matrix form (as in any quantum mechanics book) a simple calculation shows that (a a+ - a+a) is unit matrix 1 with infinite dimension, then [ x , p ] = i1  and its trace will tend to infinite, but interestingly if you calculate the trace of  a a+ and a+a separately you will find them equal to a sum of natural number series i.e. (1+2+3+....) , then Tr( a a+) = Tr( a+a) !! this may only occur where we use the infinite dimensional space representations such as Hilbert space, where trace is a sum of  infinite series that in spite of the validity of Tr( a a+) = Tr( a+a)   we have [ x , p ] = i1 . :)

Dear Hrushikesh, Oleksandr has already answered your question, so that here I restrict myself to two general remarks. Firstly, considering a matrix representation of the operators x and p, the commutator of these two matrices is proportional to the unit matrix. Clearly, the trace of this matrix is unbounded for the size of this matrix approaching infinity. Your argument, if correct, would be tantamount to proving the equality 0=∞. Secondly, historically Luttinger [1] produced an incorrect solution of a one-dimensional model of interacting particles named after him by neglecting the unbounded spectrum (from below) of this model. Two years later, in 1965, Mattis and Lieb [2] corrected Luttinger's trivial solution by introducing an appropriate cut off on the spectrum of the model.

[1] J.M. Luttinger, J. Math. Phys. 4, 1154 (1963). 

[2] D.C. Mattis, and E.H. Lieb, J. Math. Phys. 6, 304 (1965). 

The dimension of X and P operators is ∞. and so on their Trace. When you calculate :

Trace (XP) - Trace (PX) = -i hbar Trace(I)

∞ - ∞ = ∞

which is not a wrong mathematical expression. The R.H.S ∞ is because the identity matrix, I, 's dimension are ∞ too.

In other word let Trace (XP) = Trace (PX) = A you may conclude from A-A=/ 0 that A should be infinity.

Dear Hrushikesh, Oleksandr has already answered your question,and the Sahar Arabzadeh make a example for you.If you have  two operators A and B are bounded. The relation trace(AB)=trace(BA) is right.If the two operators A and B are not bounded,we did not can defint trace of the operator.

Dear Yonghong, it sounds that here is some misunderstanding about the trace of matrix representation of an operator and "bounded" and "unbounded" operators in this thread. The bounded and unbounded operators indicates a precise properties of operators in a Banach space. However if you equate the bounded operators with finite dimensional matrix representation, then for a well known counter example of your statement, recall the definition of "Density matrix" operator in a Hilbert space, i.e. " a Hermitian matrix (possibly  infinite dimensional)  with trace 1".Therefore not only trace calculation of infinite dimensional matrices is possible but also there are bounded operators with infinite dimensional matrix representation.

Dear Manouchehr Amiri,as you say ,but Trace[X,P]=ihbar*Trace(I)≠Trace (XP) - Trace (PX) = 0,can you give me a advance?

@ Yonghong,If you follow my first post and work out the matrix form of  a+a and aa+ you will arrive to this result that in spite of the Tr( a a+) = Tr( a+a) we have the equation  [ a , a+] = 1 .