On a hunch, I decided to check whether 729 is a 6th power, and found that it is:
729^(1/6) = 3.
Next, I decided to divide x-3 into x^6 + 0x^5 - 9x^4 + 0x^3 - 81x^2 + 0x + 729, through synthetic division and using 3 as my divisor. This left no remainder. The coefficients of the quotient were as follows: 3 3 0 0 -81 -243, which represents:
3x^5 + 3x^4 + 0x^3 + 0x^2 -81x -243
Next, I applied "factoring by grouping:"
3x^4(x+1) -81(x+3) = (x+3)(3x^4 - 81).
Note that 3x^4 - 81 is the same as 3(x^4-27).
Thus, the original polynomial factors into (x-3)(3)(x^4 - 27).