How do you factor (1-x^9)(1−x9)?

1 Answer
George C.
Nov 7, 2015

You can use the difference of cubes identity twice to partially factor this as:

1-x^9 = (1-x)(1+x+x^2)(1+x^3+x^6)1−x9=(1−x)(1+x+x2)(1+x3+x6)

or more thoroughly to find:

1-x^9 = (1-x)(1+x+x^2)(1-2 cos((2pi)/9)x + x^2)(1-2 cos((4pi)/9)x+x^2)(1-2 cos((8pi)/9)x+x^2)1−x9=(1−x)(1+x+x2)(1−2cos(2π9)x+x2)(1−2cos(4π9)x+x2)(1−2cos(8π9)x+x2)

Explanation:

The difference of cubes identity is:

a^3-b^3 = (a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2)

We can use this identity to partially factor 1-x^91−x9 as follows:

1-x^9 = 1^3-(x^3)^3 = (1-x^3)(1+x^3+(x^3)^2))1−x9=13−(x3)3=(1−x3)(1+x3+(x3)2))

=(1^3-x^3)(1+x^3+x^6)=(13−x3)(1+x3+x6)

= (1-x)(1+x+x^2)(1+x^3+x^6)=(1−x)(1+x+x2)(1+x3+x6)

It is possible to factor 1+x^3+x^61+x3+x6 further into quadratics with Real coefficients, expressible in terms of coscos:

1+x^3+x^6 = (1-2 cos((2pi)/9)x + x^2)(1-2 cos((4pi)/9)x+x^2)(1-2 cos((8pi)/9)x+x^2)1+x3+x6=(1−2cos(2π9)x+x2)(1−2cos(4π9)x+x2)(1−2cos(8π9)x+x2)

To understand how this works, note that De Moivre's Theorem gives us:

(cos x + i sin x)^n = cos(nx) + i sin(nx)(cosx+isinx)n=cos(nx)+isin(nx)

Putting n = 9n=9 and x = (2k pi)/9x=2kπ9, where k in ZZ we find:

(cos ((2k pi)/9) + i sin ((2k pi)/9))^9 = 1

This gives us 9 distinct 9th roots of 1.

Every third one of these roots is a cube root of 1, so a zero of 1-x^3 = (1-x)(1+x+x^2). The others are zeros of the remaining factor (1+x^3+x^6) These 6 remaining zeros occur in Complex conjugate pairs which combine to give factors with Real coefficients.

For example:

(x-cos((2pi)/9)-i sin((2pi)/9))(x-cos((16pi)/9)-i sin((16pi)/9))

=(x-cos((2pi)/9)-i sin((2pi)/9))(x-cos((2pi)/9)+i sin((2pi)/9))

=x^2-2cos((2pi)/9)x+(cos^2((2pi)/9) + sin^2((2pi)/9))

=x^2-2cos((2pi)/9)x+1

Similarly for (4pi)/9 and (8pi)/9

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