How do you factor #8000 + y^3#?

1 Answer
Jan 1, 2016

Answer:

Use the sum of cubes identity to find:

#8000+y^3=(20+y)(400-20y+y^2)#

Explanation:

Both #8000 = 20^3# and #y^3# are perfect cubes.

Use the sum of cubes identity:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

with #a=20# and #b=y# as follows:

#8000+y^3#

#=20^3+y^3#

#=(20+y)(20^2-20y+y^2)#

#=(20+y)(400-20y+y^2)#