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:
![\sqrt{ - 25} = \sqrt{25} \: \times \sqrt{ - 1} \\ \sqrt{25} \: \times i = 5i](https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%2025%7D%20%20%3D%20%20%5Csqrt%7B25%7D%20%20%5C%3A%20%20%5Ctimes%20%20%5Csqrt%7B%20-%201%7D%20%20%5C%5C%20%20%5Csqrt%7B25%7D%20%20%20%5C%3A%20%5Ctimes%20i%20%3D%205i)
So in plain terms, it's as simple as this:
![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](https://tex.z-dn.net/?f=i%20%3D%20%20%5Csqrt%7B%20-%201%7D%20%20%5C%5C%20%20%7Bi%7D%5E%7B2%7D%20%20%3D%20%20%5Csqrt%7B%20-%201%7D%20%20%5Ctimes%20%20%5Csqrt%7B%20-%201%7D%20%20%3D%20%20-%201%20%5C%5C%20%20%7Bi%7D%5E%7B3%7D%20%20%3D%20%5Csqrt%7B%20-%201%7D%20%5Ctimes%20%5Csqrt%7B%20-%201%7D%20%5Ctimes%20%5Csqrt%7B%20-%201%7D%20%5C%5C%20%20%7Bi%7D%5E%7B3%7D%20%20%3D%20%20-%201%5C%3A%20%20%5Ctimes%20%5Csqrt%7B%20-%201%7D%20%3D%20%20-%20%5Csqrt%7B%20-%201%7D%20%3D%20%20-%20i)
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.
![{i}^{even \: } = - 1 \: or \: 1 \: (depends) \\ {i}^{odd} = - i \: or \: i = \: alternates \\ between \: + and \: - \: i](https://tex.z-dn.net/?f=%20%7Bi%7D%5E%7Beven%20%5C%3A%20%7D%20%20%3D%20%20-%201%20%5C%3A%20or%20%5C%3A%201%20%5C%3A%20%28depends%29%20%5C%5C%20%20%7Bi%7D%5E%7Bodd%7D%20%20%3D%20-%20i%20%5C%3A%20%20or%20%5C%3A%20i%20%20%3D%20%20%5C%3A%20alternates%20%5C%5C%20between%20%5C%3A%20%20%2B%20and%20%20%5C%3A%20%20-%20%20%5C%3A%20i)
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!!
![{i}^{6} \: and \: 6 \: is \: even \:and \: not \: divisible \\ by \: 4 \: therefore \: our \: rule \: is \\ {i}^{6} = - 1](https://tex.z-dn.net/?f=%20%7Bi%7D%5E%7B6%7D%20%20%5C%3A%20and%20%5C%3A%206%20%5C%3A%20is%20%5C%3A%20even%20%5C%3Aand%20%5C%3A%20not%20%5C%3A%20divisible%20%20%5C%5C%20%20by%20%5C%3A%204%20%5C%3A%20%20therefore%20%20%5C%3A%20our%20%20%5C%3A%20rule%20%5C%3A%20is%20%5C%5C%20%20%7Bi%7D%5E%7B6%7D%20%20%3D%20%20-%201)
Proof: