How to solve the equation x^8 - 64 = 0?

Let's recall first the formula for the difference of
squares:

a^2 - b^2
=(a-b)(a+b)

Now,we'll put a^2 = x^8 = (x^4)^2 and b^2 = 64
= 8^2

We'll re-write the given equation, emphasizing on the
difference of squares:

 (x^4)^2 - 8^2 = (x^4 - 8)(x^4 +
8)

x^4 - 8 is also a difference o squares, whose terms are:
a = x^2 and b = 2sqrt2

(x^4 - 8)(x^4 + 8) = (x^2 -
2sqrt2)(x^2 + 2sqrt2)(x^4 + 8)

Now, we'll solve the
equation:

x^8 - 64 = 0

x^8 -
64 = (x^2 - 2sqrt2)(x^2 + 2sqrt2)(x^4 + 8)

(x^2 -
2sqrt2)(x^2 + 2sqrt2)(x^4 + 8) = 0

x^2 - 2sqrt2 =
0

x^2 = 2sqrt2

x1 =
+sqrt(2sqrt2)

x2 =
-sqrt(2sqrt2)

x^2 + 2sqrt2 =
0

x3 = +i*sqrt(2sqrt2)

x4 =
-i*sqrt(2sqrt2)

The only real roots of the
equation are {+sqrt(2sqrt2) ; -sqrt(2sqrt2)}.