How to solve this polynomials problem

(2k^3/3k^-2)^-2

Please explain how to do this to me along with the answer!

To solve this polynomial you first distribute the -2 exponent into the parenthesis.
You would get: (2^-2k^-6/3^-2k^4)
Next, because negative exponents mean you flip the numbers, you would put everything negative on the bottom to the top and everything negative from the top to the bottom and leave positives where they are.
You would get: (3^2/ 2^2k^6k^4)
Next you will combine like terms and apply the exponents to coefficients 2 and 3 and simplify if possible.

You would be left with the final answer of:
9/4k^10

First, I'd factor out the two constant coefficients (2 and 3) --->

(2k^3/3k^-2)^-2 = [(2/3)k^3/k^-2]^-2

Note that k^3/k^-2 = (k^3 k^2) = k^5, then

[(2/3)k^5]^-2 = (9/4)k^-10 = 9/(4k^10)

Note: This is not a "polynomial problem"; it's merely how to deal with exponents.