(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)
Here is something that may clue you in......
Factoring a Difference of Cubes:
a3 – b3 = (a – b)(a2 + ab + b2)
If you are asked to factorize it, that is a good start....
((1/2x)^3 - 2^3 ) can you do it now?
(1/2x)^3 - 2^3 )
answer: (2x-2)(4x^2 + 4x+4) Correct?
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