Completely factor: 8m3 + 64r3

Question

Completely factor: 8m3 + 64r3

a. (2m - 4r)(4m2 + 8mr + 16r2)

b. (2m + 4r)(4m2 + 8mr + 16r2)

 

c. (2m + 4r)(4m2 - 8mr + 16r2)

d. (2m - 4r)(4m2 - 8mr + 16r2)

Answer 1

8m³ + 64r³ = (2m)³ + (4r)³

 

This is the sum of 2 cubes.  The formula for the sum of 2 cubes is:

a³ + b³ = (a+b)(a² - ab + b²)

In this case a = 2m, b = 4r

 

8m³ + 64r³ = (2m + 4r)[(2m)² - (2m)(4r) + (4r)²]

 

= (2m + 4r)(4m² - 8mr + 16r²)

 

This is answer c

 

However, this could actually be factored further since we have all even coefficients:

 

= 2(m+2r)(4)(m² - 2mr + 4r²)

= 8(m+2r)(m² - 2mr + 4r²)

Answer 2

This is a sum of cubes.

 

8m³ + 64r³ = (2m)³ + (4r)³ = (2m + 4r) [(2m)² - (2m)(4r) + (4r)²]

 

Answer (c) is correct, but you can also factor out a 2 from (2m - 4r).