I guess the question was on expanding the expression rather than factorizing as some people are talking above.
There is a different way to view the calcuation of the expansion in my opinion.
The spirit is to use a similar combinatorics idea of binomial theorem for (x+b)^n, except now b is changing along the way.
Then the expansion could be viewed as a matter of combinatorics.
However, because the difference in b in each terms, the combinatorics doesn't come out nicely like the binomial case. Rather, it's a sum of terms that are having or missing some of the factors between 1 to 10.
[e.g. the x^9 term would be 1+...+10,
x^8 terms is the sum of the product of the 2 number combinations.
etc...and technically you would only need to work till the x^5 term, becuase of the symmetrical nature, the other terms would be sum of (10!/product of combinations)]
But I am not saying this is necessarily the faster way to calculate this, since the summation of combinations is still quite involved.
