How do you factor $1331x^3 – 8y^3$ ?

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Answer 1 · 2016-01-07T23:57:01

Use the difference of cubes identity to find:

$1331x^3-8y^3 =(11x-2y)(121x^2+22xy+4y^2)$

Both $1331x^3 = (11x)^3$ and $8y^3 = (2y)^3$ are perfect cubes.

Use the difference of cubes identity, which can be written:

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

with $a=11x$ and $b=2y$ as follows:

$1331x^3-8y^3$

$=(11x)^3-(2y)^3$

$=(11x-2y)((11x)^2+(11x)(2y)+(2y)^2)$

$=(11x-2y)(121x^2+22xy+4y^2)$

The remaining quadratic factor cannot be factored into linear factors with Real coefficients (as you can tell by checking its discriminant).

You can factor it using Complex coefficients:

$=(11x-2y)(11x-2omega y)(11x-2omega^2 y)$

where $omega = -1/2+sqrt(3)/2 i$ is the primitive Complex cube root of $1$ .

How do you factor 1331x^3 – 8y^31331x^3 – 8y^3?