The simplified version of "square root of m^17" can be written as m^(17/2). Since 17 is an odd number, it cannot be divided evenly and instead we can only simplify out the closest even number, which is 16. The square root of m^16 is m^8, leaving a singular m on the inside of the radical.
From there, we just need to square the 3 on the outside of the radical in order to put in back inside, and 3^2 is 9. 9 times the 7 that is left inside the radical = 63, so the whole problem can be written as follows:
