# How do you factor #x^9 - 27#?

##### 1 Answer

#### Answer:

You can use the difference of cubes identity to help factor this, but to find all the factors with Real coefficients requires a little more...

#### Explanation:

You can partially factor

The difference of cubes identity can be written:

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

In our case we can put

#x^9-27 = (x^3)^3-3^3 = (x^3-3)((x^3)^2+(x^3)(3)+3^2)#

#=(x^3-3)(x^6+3x^3+9)#

The remaining cubic factor can itself be treated as a difference of cubes:

#x^3-3 = x^2-(root(3)(3))^3 = (x-root(3)(3))(x^2+root(3)(3)x+root(3)(3)^2)#

The remaining sextic factor can be factored as the product of three quadratics:

#x^6+3x^3+9 =#

#(x^2-2cos((2pi)/9)root(3)(3)x+root(3)(3)^2) * #

#(x^2-2cos((4pi)/9)root(3)(3)x+root(3)(3)^2) * #

#(x^2-2cos((8pi)/9)root(3)(3)x+root(3)(3)^2)#

To see why requires some Complex arithmetic.

The primitive Complex

#alpha = cos((2pi)/9) + i sin((2pi)/9)#

Then all the

Apart from

The other

#(x-alpha root(3)(3))(x-alpha^8 root(3)(3)) = x^2-2 cos((2pi)/9)root(3)(3)x + root(3)(3)^2#

#(x-alpha^2 root(3)(3))(x-alpha^7 root(3)(3)) = x^2-2 cos((4pi)/9)root(3)(3)x + root(3)(3)^2#

#(x-alpha^4 root(3)(3))(x-alpha^5 root(3)(3)) = x^2-2 cos((8pi)/9)root(3)(3)x + root(3)(3)^2#