It is helpful to learn the cubes of the first few integers, so you can recognize them when you see them.
1^3 = 1*1*1 = 1
2^3 = 2*2*2 = 8
3^3 = 3*3*3 = 27
4^3 = 4*4*4 = 64
5^3 = 5*5*5 = 125
Knowing these, you recognize immediately that the given binomial is the difference of two cubes. The factorization of this is a "standard form" you are expected to learn, too. x^3 - y^3 = (x-y)(x^2 + xy + y^2)
So
27a^3 - 125
= (3a)^3 - 5^3
= (3a-5)((3a)^2 + (3a)(5) + (5)^2)
= (3a - 5)(9a^2 + 15a + 25)
1^3 = 1*1*1 = 1
2^3 = 2*2*2 = 8
3^3 = 3*3*3 = 27
4^3 = 4*4*4 = 64
5^3 = 5*5*5 = 125
Knowing these, you recognize immediately that the given binomial is the difference of two cubes. The factorization of this is a "standard form" you are expected to learn, too. x^3 - y^3 = (x-y)(x^2 + xy + y^2)
So
27a^3 - 125
= (3a)^3 - 5^3
= (3a-5)((3a)^2 + (3a)(5) + (5)^2)
= (3a - 5)(9a^2 + 15a + 25)