(1/8 X^3 - 8) -Factorize.

Why don't we take the 1/8 as a common factor?

so it is:

1/8 (x^3 - 64)

Then factorize:

1/8 (x-4) (x^2 +4x +16)

^^My solution..

Answer book solution:

He Cube root both terms so that

1/8 X^3 -8 (Cube root) = (1/2 X -2) (1/4 +X + 4)

Guest Apr 15, 2019

#1**0 **

Here is something that may clue you in......

Factoring a Difference of Cubes:

a3 – b3 = (a – b)(a2 + ab + b2)

ElectricPavlov Apr 15, 2019

#3**0 **

If you are asked to factorize it, that is a good start....

((1/2x)^3 - 2^3 ) can you do it now?

ElectricPavlov
Apr 15, 2019

#4

#5**0 **

(1/2x)^3 - 2^3 )

answer: (2x-2)(4x^2 + 4x+4) Correct?

Guest Apr 15, 2019

edited by
Guest
Apr 15, 2019

#7**+2 **

Your answer is just as valid as the book's

You get

(1/8)(x - 4) ( x^2 + 4x + 16) which is actually the same as

(1/2)(1/4)(x - 4) ( x^2 + 4x + 16) =

(1/2)(x - 4) (1/4) ( x^2 + 4x + 16) =

(1/2 x - 2) (x^2/4 + x + 4) ⇒ "book answer"

Yours is actually easy to work with .....it avoids the nasty fractions in the terms

CPhill Apr 15, 2019