How do you factor: $y= x^3 - 27$ ?

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Answer 1 · 2016-05-05T19:03:02

$y=x^3-27=(x-3)(x^2+3x+9)$

Both $x^3$ and $27=3^3$ are perfect cubes. So it is natural to use the difference of cubes identity:

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

with $a=x$ and $b=3$ as follows:

$x^3-27$

$=x^3-3^3$

$=(x-3)(x^2+(x*3)+3^2)$

$=(x-3)(x^2+3x+9)$

How do you factor: y= x^3 - 27 y= x^3 - 27 ?