factor completely

1) 8x^3 + 27y^3

2) u^3 + 125v^3

Note that

1) 8x^3+27y^3 ≡ (2x)^3+(3y)^3
and apply the following identity:
a^3+b^3 = (a+b)(a^2-ab+b^2)
hint: substitute a=2x, b=3y
2) same strategy as the first.

To factor completely, you need to find the factors of the given expressions. Let's factor each expression separately:

1) To factor 8x^3 + 27y^3, we can use the sum of cubes formula. The sum of cubes formula states that a^3 + b^3 can be factored as (a + b)(a^2 - ab + b^2).

In this case, a = 2x and b = 3y. So we have:

8x^3 + 27y^3 = (2x + 3y)((2x)^2 - (2x)(3y) + (3y)^2)
= (2x + 3y)(4x^2 - 6xy + 9y^2)

Therefore, the expression 8x^3 + 27y^3 can be completely factored as (2x + 3y)(4x^2 - 6xy + 9y^2).

2) To factor u^3 + 125v^3, we can use the sum of cubes formula again. In this case, a = u and b = 5v. So we have:

u^3 + 125v^3 = (u + 5v)(u^2 - u(5v) + (5v)^2)
= (u + 5v)(u^2 - 5uv + 25v^2)

Therefore, the expression u^3 + 125v^3 can be completely factored as (u + 5v)(u^2 - 5uv + 25v^2).