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

1 Answer
Nov 16, 2015

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 #x^9-27# by first recognising that it is a difference of cubes.

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 #a=x^3# and #b=3# to find:

#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 #9#th root of #1# is:

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

Then all the #9#th roots of #1# are #alpha, alpha^2,..., alpha^9 = 1#

Apart from #alpha^9 = 1#, these root occur in Complex conjugate pairs, which we can combine to find quadratic factors with Real coefficients. The roots #alpha^3#, #alpha^6# and #alpha^9=1# are all cube roots of #1# relating to the factors of #x^3-3#.

The other #6# roots combine in pairs to give the three other quadratic factors of the sextic:

#(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#