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# Factorization question

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(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..

He Cube root both terms so that

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

Apr 15, 2019

#1
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Here is something that may clue you in......

Factoring a Difference of Cubes:

a3 – b3 = (a – b)(a2 + ab + b2)

Apr 15, 2019
#2
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Ohh!! so in any difference of cubes, it must be solved in that form first?

Guest Apr 15, 2019
#3
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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
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Thank you so much got it!!

Apr 15, 2019
#5
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(1/2x)^3 - 2^3 )

Guest Apr 15, 2019
edited by Guest  Apr 15, 2019
#6
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a = 1/2x    b = 2

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

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

ElectricPavlov  Apr 15, 2019
#7
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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   Apr 15, 2019