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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..

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) 

 Apr 15, 2019
 #1
avatar+19832 
+2

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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+1

Ohh!! so in any difference of cubes, it must be solved in that form first?

Guest Apr 15, 2019
 #3
avatar+19832 
+2

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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0

Thank you so much got it!!

 Apr 15, 2019
 #5
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0

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

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

Guest Apr 15, 2019
edited by Guest  Apr 15, 2019
 #6
avatar+19832 
0

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
avatar+104937 
+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

 

cool cool cool

 Apr 15, 2019

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