Factorise..x^6 -729 y^6
This looks like it could be a difference of squares (x^6-y^6), but 729 = 9^3, so we'll have to treat it as a difference of cubes.
(x^2)^3 - (9y^2)^3 = (x-9y)(x^4+9y^2+81y^4)
actually, in thinking about it, 729 = 9^3 = 3^6 = 27^2. So,
x^6-729 = (x^3)^2 - (27y^3)^2
= (x^3-27y^3)(x^3+27y^3)
= (x-3y)(x^2+3xy+9y^2)(x+3y)(x^2-3xy+9y^2)
To factorize the expression x^6 - 729y^6, we first recognize that it is a difference of two cubes. The expression can be written as:
(x^2)^3 - (9y^2)^3
The difference of two cubes can be factorized using the formula:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Applying this formula to our expression, we have:
(x^2 - 9y^2)((x^2)^2 + (x^2)(9y^2) + (9y^2)^2)
(x^2 - 9y^2)(x^4 + 9x^2y^2 + 81y^4)
Therefore, the factorization of x^6 - 729y^6 is (x^2 - 9y^2)(x^4 + 9x^2y^2 + 81y^4).