Factorise: x^6} - 729.

We are given the expression:

x^6 - 729

We recognize that 729 is a perfect cube:

729 = 9^3 = (3^2)^3 = 27^2

So the expression becomes:

x^6 - 729 = x^6 - 3^6

This is a difference of squares, which can be factored using the formula:

a^2 - b^2 = (a - b)(a + b)

In our case, a = x^3 and b = 3^3:

(x^3)^2 - (3^3)^2

So, the expression factors as:

(x^3 - 27)(x^3 + 27)

Next, we recognize that both x^3 - 27 and x^3 + 27 are difference and sum of cubes. We can further factor these:

x^3 - 27 is a difference of cubes, which can be factored using the formula:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Here, a = x and b = 3, so we get:

x^3 - 27 = (x - 3)(x^2 + 3x + 9)

x^3 + 27 is a sum of cubes, which can be factored using the formula:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)

Here, a = x and b = 3, so we get:

x^3 + 27 = (x + 3)(x^2 - 3x + 9)

Thus, the complete factorization of the expression x^6 - 729 is:

(x - 3)(x^2 + 3x + 9)(x + 3)(x^2 - 3x + 9)