Question 71067
Factor completely: 27x^6 - 8y^9
:
Factor as the difference of cubes: (x^3 - y^3) = (x - y)(x^2 + xy + y^2)
:
That would be: (27x^6 - 8y^9) = (3x^2 - 2y^3) (9x^4 + 3x^2*2y^3 + 4y^6)
:
Or: (27x^6 - 8y^9) = (3x^2 - 2y^3) (9x^4 + 6x^2y^3 + 4y^6)