Posted by **amy** on Friday, November 12, 2010 at 3:18pm.

What is the fundamental theorem of algebra for x^6-64?

- College algebra -
**Reiny**, Friday, November 12, 2010 at 4:25pm
I would understand it to say that

x^6 - 64 = 0 has 6 roots

Proof:

first treat x^6 as a difference of squares

x^6 - 64

= (x^3 - 8)(x^3 + 8)

now as a difference of cubes for the first factor, and a sum of cubes for the second part

= (x-2)(x^2 + 2x + 4)(x+2)(x^2 - 2x + 4)

so x^6 - 64 = 0

(x-2)(x^2 + 2x + 4)(x+2)(x^2 - 2x + 4) = 0

x-2=0 ----> x=2

x+2=0 ----> x=-2

x^2+2x+4)=0 ---> x = (-2 ± √-12)/2 = -1 ± √3 i

x^2 - 2x + 4=0 --> x = (2 ± √-12)/2 = 1 ± √3 i

6 roots!

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