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:

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

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