how do i factorization of 64x³ + 8

Question

explain factorization of 64x³ + 8

Answer 1

 

The GENERAL pattern is:

A^3 + B^3 = (A +B) (A^2  -  AB +  B^2)

 

In this case A = 4x and y = 2 because 64x^3 = (4x)(4x)(4x) and 8 = 2*2*2 are their respective cubes.

 

The factorization is then

 

(4x +2   ) (  16x^2 - 8x + 4  )

 

Finally a factor of 2 can be factored out of the first binomial and

a factor of 4 can be factored out of the trinomial.

 

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

 

CHeck by multiplying them:

8 ( 8x^3 - 4x^2 + 2x + 4x^2 - 2x + 1)

8( 8x^3  + 1)

64x^3 + 8

 

Yes, you can factor out the 8 first.

64x^3 + 8 = 8 ( 8x^3 + 1)

 

This time A = 2x and B = 1 with respect to the pattern, as 8x^3 = (2x)(2x)(2x) and 1= 1*1*1

 

The factorization is 8 ( 2x +1   )( 4x^2 - 2x + 1)

Answer 2

64x³ + 8 = 8(8x³ + 1) = 8[(2x)³ + 1]

 

Note that 

 

(2x)³ + 1

 

is a sum of cubes.  Can you apply the factorization of a sum of cubes and finish from here?