How would you solve this in a test

I was wondering how you would show your working out when solving:
X^6=64

Thanks

x^6=64
(x^3)^2 = 64
x^3 = ±8
x = ±2

Seems odd that people would miss an obvious second solution. If you want to show thorough work, do the following:

x^6 = 64 ==> x^6 - 64 = 0

Factor as a difference of squares, and use sum and difference of cubes.

(x^3 - 8)(x^3 + 8) = 0 ==> (x - 2)(x² + 2x + 4)(x + 2)(x² - 2x + 4) = 0.

The quadratics are prime factors. So x = 2 or x = -2. (These are the two real solutions.)

Think about what's being said. It's saying that some number, times itself six times equals 64. So...

x*x*x*x*x*x=64
x^2*x^2*x^2=64, combine x*x=x^2, there are three
(x^2)^3=64, simplify since (x^a)^b=x^(a*b)
x^2=64^(1/3), cubed root both sides
x^2=4, simplify
x=sqrt(4)=2, finally the answer

x^6=64
sq. root^6(x^6)=sq. root^6(64)
x=2

x=(64)^(1/6)

x=2