How do you factor #24x^6 – 1029y^3#?
1 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