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

##### 1 Answer

#### Answer:

Use the difference of cubes identity to find:

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

#### Explanation:

First separate out the common scalar factor

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

Next notice that both

The difference of cubes identity can be written:

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

Use this with

#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