How do you factor #1000x^3 + 216 #?

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Answer 1 · 2015-12-31T11:30:19

Use the sum of cubes identity to get:

#1000x^3+216#

#=(10x+6)(100x^2-60x+36)#

#=8(5x+3)(25x^2-15x+9)#

Both #1000x^3 = (10x)^3# and #216=6^3# are perfect cubes, so we can use the sum of cubes identity:

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

with #a=10x# and #b=6# as follows:

#1000x^3+216#

#=(10x)^3+6^3#

#=(10x+6)((10x)^2-(10x)(6)+6^2#

#=(10x+6)(100x^2-60x+36)#

Alternatively, separate out the common scalar factor #8#, to get:

#=8(5x+3)(25x^2-15x+9)#