How do you factor $24x^6 – 1029y^3$ ?

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Answer 1 · 2016-01-02T14:13:43

Use the difference of cubes identity to find:

$24x^6-1029y^3=3(2x^2-7y)(4x^4+14x^2y+49y^2)$

First separate out the common scalar factor $3$ :

$24x^6-1029y^3=3(8x^6-343y^3)$

Next notice that both $8x^6 = (2x^2)^3$ and $343y^3 = (7y)^3$ are perfect cubes.

The difference of cubes identity can be written:

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

Use this with $a=2x^2$ and $b=7y$ to find:

$8x^6-343y^3$

$=(2x^2)^3-(7y)^3$

$=(2x^2-7y)((2x^2)^2+(2x^2)(7y)+(7y)^2)$

$=(2x^2-7y)(4x^4+14x^2y+49y^2)$

Putting it all together:

$24x^6-1029y^3=3(2x^2-7y)(4x^4+14x^2y+49y^2)$

This is as far as we can go with Real coefficients.

If we allow Complex coefficients then this can be factored a little further:

$=3(2x^2-7y)(2x^2-7omegay)(2x^2-7omega^2y)$

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

How do you factor 24x^6 – 1029y^324x^6 – 1029y^3?