Please explain how to do this?i^6=A. -1B. 1C. -iD. i

Ok, so "i" was invented to allow algebra math problems to continue, despite containing the square root of a negative. The square root of a negative can't be solved because anything squared will become positive.
In other words, square root of -25 is not solvable because -x * -x = +x^2
To make solving manageable, mathematicians created the square root of negative one (-1) as this imaginary entity (i).
Let's use that square root of -25 again:
[tex] \sqrt{ - 25} = \sqrt{25} \: \times \sqrt{ - 1} \\ \sqrt{25} \: \times i = 5i[/tex]

So in plain terms, it's as simple as this:
[tex]i = \sqrt{ - 1} \\ {i}^{2} = \sqrt{ - 1} \times \sqrt{ - 1} = - 1 \\ {i}^{3} = \sqrt{ - 1} \times \sqrt{ - 1} \times \sqrt{ - 1} \\ {i}^{3} = - 1\: \times \sqrt{ - 1} = - \sqrt{ - 1} = - i[/tex]
without doing too many examples, I want you to understand this pattern:
whenever the exponent of i is EVEN, then the answer will not have a radical !!
Why? because every couplet of i's will = i^2 which = -1
BUT an ODD numbered exponent of i will always leave that extra i after all the couplets become -1.
[tex] {i}^{even \: } = - 1 \: or \: 1 \: (depends) \\ {i}^{odd} = - i \: or \: i = \: alternates \\ between \: + and \: - \: i[/tex]
It's a little complicated in that every couplet will become -1, but if you have pairs (2) of COUPLETS, then -1×-1 = 1
So a couplet (2) × a pair (2) of couplets = 2×2=4
That means that if the even exponent is divisible by 4 (4, 8, 12, 16, etc.), then the answer will ALWAYS be +1. Otherwise the even exponent 2, 6, 10, 14, etc.) will result in a -1.

Now for our actual problem!!
[tex] {i}^{6} \: and \: 6 \: is \: even \:and \: not \: divisible \\ by \: 4 \: therefore \: our \: rule \: is \\ {i}^{6} = - 1[/tex]
Proof:
[tex] {i}^{6} = ( \sqrt{ - 1} ) \times ( \sqrt{ - 1} ) \times ( \sqrt{ - 1} ) \\ \times ( \sqrt{ - 1} ) \times ( \sqrt{ - 1} ) \times ( \sqrt{ - 1} ) \\ = ( - 1) \times ( - 1) \times ( - 1) \\ = ( + 1) \times ( - 1) = - 1[/tex]