What jumps out at me is that everything in this expression have the right to be express in regards to cubes:

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

and, in fact, us can factor this utilizing the basic formula:

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

so let"s perform that.

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


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If we permit irrational coefficients, climate this sextic expression will aspect as far as a mixture that linear and quadratic factors.

Use the complying with 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 the 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 distinction 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 every 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 difficulty is that the coefficients #8=2^3# and also #27=3^3# naturally allude to treating this as a difference of cubes an initial to find:

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

If us then decide to permit irrational coefficients climate the very 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 2nd is not so straightforward...

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

will not element as a "quadratic in #x^2# and also #y^2#" with actual 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# through #9y^4# by putting #b=sqrt(3)y#, to find: