Question

Factor 64x^12+27y^3 completely.

Answer 1

To factor 64x^12 + 27y^3 completely, we can use the difference of cubes formula. The expression is completely factored as (4x^4 - 3y)(16x^8 + 12x^4y + 9y^2).

Step-by-step explanation:

To factor 64x^12 + 27y^3 completely, we can use the difference of cubes formula, which states that a^3 - b^3 = (a - b)(a^2 + ab + b^2). In this case, a = 4x^4 and b = 3y. So, we have:

64x^12 + 27y^3 = (4x^4)^3 - (3y)^3 = (4x^4 - 3y)((4x^4)^2 + (4x^4)(3y) + (3y)^2) = (4x^4 - 3y)(16x^8 + 12x^4y + 9y^2)

Therefore, the expression is completely factored as (4x^4 - 3y)(16x^8 + 12x^4y + 9y^2).

Answer 2

64 is 4^3
x^12 is (x^4)^3
27 is 3^3
using the formula (a^3+b^3)=a^2-ab+b^2), we have
(4x^4+3y)(16x^8-12x^4y+9y^2)