The shortcut method for doing this problem is to think about seven goes into 20 how many full times, and the full times go on the outside. So seven goes into 22 full times with six leftover. And then we're going to do the same thing for the YS. We're going to think 7 goes into 31. How many full times? Seven goes into 31? Four full times with three leftover. Then we're going to think about 7 going into the 19. Seven goes into the 19 twice with five leftover. So the shortcut method gives us X ^2, y to the 4th, Z ^2, the 7th root of X to the 6th, y ^3 Z to the 5th. A longer method would be to think of writing each of those as fractions. So X to the 20 sevenths Y equal Y to the 31 seventh, Z to the 19 seventh. So 20 sevenths would be 2 + 6 sevenths, 31 sevenths would be 4 and 3 sevenths, and 19 sevenths would be two and five sevenths. So we could think of the rewriting that as X ^2 * X to the 6th sevenths, Y to the fourth times, Y to the three 7th, Z ^2 * Z to the five 7th. If we took all the whole number exponents, the X ^2, y to the 4th Z ^2 and put them together, and we took all the rest, X to the 6th sevenths, Y to the three sevenths, Z to the 5 sevenths, we realized that those fractions all have a 7th as the denominator. So we'd have X ^2, y to the 4th, Z ^2 the 4th root of X to the 6th.