How do you factor #8x^6 - 27y^6#?

2 Answers

Answer:

#8x^6-27y^6=2^3(x^2)^3-3^3(y^2)^3=(2x^2)^3-(3y^2)^3=(2x^2-3y^2)(4x^4+6x^2y^2+9y^4)#

Explanation:

What jumps out at me is that everything in this expression can be expressed in terms of cubes:

#8x^6-27y^6=2^3(x^2)^3-3^3(y^2)^3=(2x^2)^3-(3y^2)^3#

and, in fact, we can factor this using the general formula:

#A^3-B^3=(A-B)(A^2+AB+B^2)#

so let's do that.

#(2x^2)^3-(3y^2)^3=(2x^2-3y^2)(4x^4+6x^2y^2+9y^4)#

May 4, 2017

Answer:

#8x^6-27y^6#

#= (sqrt(2)x-sqrt(3)y)(2x^2+sqrt(6)xy+3y^2)(sqrt(2)x+sqrt(3)y)(2x^2-sqrt(6)xy+3y^2)#

Explanation:

If we allow irrational coefficients, then this sextic expression will factor as far as a mixture of linear and quadratic factors.

Use the following identities:

Difference of squares:

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

Difference of cubes:

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

Sum of cubes:

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

Using the difference of squares:

#8x^6-27y^6 = (2sqrt(2)x^3)^2-(3sqrt(3)y^3)^2#

#color(white)(8x^6-27y^6) = (2sqrt(2)x^3-3sqrt(3)y^3)(2sqrt(2)x^3+3sqrt(3)y^3)#

#color(white)(8x^6-27y^6) = ((sqrt(2)x)^3-(sqrt(3)y)^3)((sqrt(2)x)^3+(sqrt(3)y)^3)#

Using the difference of cubes:

#(sqrt(2)x)^3-(sqrt(3)y)^3 = (sqrt(2)x-sqrt(3)y)((sqrt(2)x)^2+(sqrt(2)x)(sqrt(3)y)+(sqrt(3)(y))^2)#

#color(white)((sqrt(2)x)^3-(sqrt(3)y)^3) = (sqrt(2)x-sqrt(3)y)(2x^2+sqrt(6)xy+3y^2)#

Using the sum of cubes:

#(sqrt(2)x)^3+(sqrt(3)y)^3 = (sqrt(2)x+sqrt(3)y)((sqrt(2)x)^2-(sqrt(2)x)(sqrt(3)y)+(sqrt(3)(y))^2)#

#color(white)((sqrt(2)x)^3+(sqrt(3)y)^3) = (sqrt(2)x+sqrt(3)y)(2x^2-sqrt(6)xy+3y^2)#

Putting it all together:

#8x^6-27y^6#

#= (sqrt(2)x-sqrt(3)y)(2x^2+sqrt(6)xy+3y^2)(sqrt(2)x+sqrt(3)y)(2x^2-sqrt(6)xy+3y^2)#

#color(white)()#
Notes

What is interesting about this problem is that the coefficients #8=2^3# and #27=3^3# naturally point to treating this as a difference of cubes first to find:

#8x^6-27y^6 = (2x^2-3y^2)(4x^4+6x^2y^2+9y^4)#

If we then decide to allow irrational coefficients then the first of these factors fairly straightforwardly as:

#2x^2-3y^2 = (sqrt(2)x-sqrt(3)y)(sqrt(2)x+sqrt(3)y)#

but the second is not so straightforward...

#4x^4+6x^2y^2+9y^4#

will not factor as a "quadratic in #x^2# and #y^2#" with real coefficients.

To factor it, we can consider the following:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

We can match #a^4# with #4x^4# by putting #a=sqrt(2)x#, #b^4# with #9y^4# by putting #b=sqrt(3)y#, to find:

#(2x^2-sqrt(6)kxy+3y^2)(2x^2+sqrt(6)kxy+3y^2)#

#= 4x^4+(12-6k^2)x^2y^2+9y^4#

So to match #12-6k^2# with #6#, we just need #k^2=1#.

Using #k=1# we find:

#(2x^2-sqrt(6)xy+3y^2)(2x^2+sqrt(6)xy+3y^2) = 4x^4+6x^2y^2+9y^4#