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#.