Expert: Paul Klarreich - 3/9/2007

Question
how do you factorise x^6 + 64?

Answer
Questioner:   Bianca
Category:  Advanced Math

 
Subject:  complex numbers
Question:  how do you factorise x^6 + 64?
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Hi, Bianca,

This is a sum-or-difference of two cubes.  The general pattern is:

a^3 + b^3 = (a + b)(a^2 - ab + b^2),  or
a^3 - b^3 = (a - b)(a^2 + ab + b^2)  

In this case, use the first one, with a = x^2, b = 4:

x^6 + 64 = (x^2 + 4)(x^4 - 4x^2 + 16)

This gets you started.  I see that you are studying complex numbers, so I think you know how to handle the (x^2 + 4)  part.  For the second part:

x^4 - 4x^2 + 16

Treat it as if it said:

a^2 - 4a + 16

and solve this as if it were a quadratic.  (which, in fact it is)

a^2 - 4a + 16 = 0

a^2 - 4a      = - 16

a^2 - 4a + 4  = - 16 + 4

  (a - 2)^2  = - 12

   a - 2     = +- 2i sqrt(3)

   a = 2 +- 2i sqrt(3)

Now your solutions are the sqrt of those:

x = +- sqrt(2 +- sqrt(3))

Two plus-or-minus signs in that means FOUR solutions, along with your +-2i for the other part gives your six roots of the equation.

Is this too much?