Question

Factor completely x^3 + 1/8

Answer 1

Answer:

((2 x + 1) (4 x^2 - 2 x + 1))/8

Step-by-step explanation:

Factor the following:

x^3 + 1/8

Put each term in x^3 + 1/8 over the common denominator 8: x^3 + 1/8 = (8 x^3)/8 + 1/8:

(8 x^3)/8 + 1/8

(8 x^3)/8 + 1/8 = (8 x^3 + 1)/8:

(8 x^3 + 1)/8

8 x^3 + 1 = (2 x)^3 + 1^3:

((2 x)^3 + 1^3)/8

Factor the sum of two cubes. (2 x)^3 + 1^3 = (2 x + 1) ((2 x)^2 - 2 x + 1^2):

((2 x + 1) ((2 x)^2 - 2 x + 1^2))/8

1^2 = 1:

((2 x + 1) ((2 x)^2 - 2 x + 1))/8

Multiply each exponent in 2 x by 2:

((2 x + 1) (2^2 x^2 - 2 x + 1))/8

2^2 = 4:

Answer:  ((2 x + 1) (4 x^2 - 2 x + 1))/8

Answer 2

$x^3+\dfrac{1}{8}=\\ \\ =x^3+2^{-3} = \\ \\ =x^3+(2^{-1})^3 =\\ \\ = (x+2^{-1})(x^2-2^{-1}x+2^{-2}) = \\ \\ = \Big(x+\dfrac{1}{2}\Big)\Big(x^2-\dfrac{x}{2}+\dfrac{1}{4}\Big)\\ \\ \\ \boxed{a^3+b^3 = (a+b)(a^2-ab+b^2)}$