125a^3-8 Factor?

Question

Factor completely.

Answer 1

(5a)^3 - 2^3

(5a - 2)(25a^2 + 10a + 4)

Answer 2

You should use the difference of cubes for this one:
(a-b) (a^2+ab+b^2)

125a^3-8 This should be quite easy as each number and variable here have perfect cube roots.
a is the cube root of 125a^3 while b is the cube root of 8

You should arrive at this answer:
(5a-2) (25a^2+10a+4)

Answer 3

125 = 5^3

use a^3-b^3 = (a-b)(a^2 + ab + b^2)

125a^3 - 8 = (5a)^3 - 2^3 =
= (5a - 2)[(5a)^2 + 10a + 4] =
= (5a - 2)(25a^2 + 10a + 4)

Now, because, the discriminant of 25a^2 + 10a + 4, 10^2 -4*4*25 =
=100 - 400 = -300 is negative, it cannot be factorized in R, moving to Z:

25a^2 + 10a + 4 has 2 zeros in Z:

a1,2 = [-10 +- sqrt(300)]/50 = [-10 +- 10sqrt(3)]/50

a1 = -1/5 + i sqrt(3)/5
a2 = -1/5 - i sqrt(3)/5

(25a^2 + 10a + 4) =
= 25[a + (1/5 - i sqrt(3)/5)] * [a + (1/5 + i sqrt(3)/5] =
= [5a + (1 - i sqrt(3))][5a + (1 + i sqrt(3))]

125a^3-8 = (5a - 2) * [5a + (1 - i sqrt(3))] *
* [5a + (1 + i sqrt(3))]

Answer 4

Yeah, difference of cubes. You have to follow this formula:

(a^3 - b^3) = (a - b)(a^2 + ab + b^2)

So,
(125a^3 - 8) =
[(5a)^3 - (2)^3] =
(5a - 2)(25a^2 + 10a + 4)

Answer 5

Difference of cubes - use the formula
(a-b) (a^2+ab+b^2)

In your example,
'a' in the formula is 5a^3
'b' in the formula is 2

Your factorization then becomes
(5a – 2)(25a² + 10a + 4)