Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Simplifying by taking roots in the numerator and denominator, assume all variables represent positive numbers, which means no absolute value brackets are necessary. So when we look at this first problem, we want to think about what is sqrt 100 / sqrt 81? Sqrt 100 is 10, sqrt 81 is 9. So our answer there is 10 / 9 or one and one ninth the next one, the cube root of 343. Sometimes it's helpful to write out some of our cubes. So if we have 1 ^3, 2 ^3, 3 ^3, 4 ^3, 5 ^3, 6 ^3, 7 ^3, yeah, let's do to 10 maybe. And yes, you're supposed to be able to do this without a calculator. 1 ^3 is 1/2 cubed is eight, 3 ^3, 27 four cubed is 64 five cubed. Is 125 six cubed. Is 216 seven cubed. Is 343, eight cubed is 512, nine cubed is 729, and 10 ^3 is 1000. So the cube root of 343 is the seven, and the cube root of 1000 is 10. So 7/10 this next 1 sqrt 25 is 5 sqrt 8 of the 5th. There's an understood 2 here, and we're going to think about two goes into that exponent OF52 full times with one leftover. The leftover stays under the radical. Two goes into 6 three full times. So there is no radical in the bottom. So five a ^2 sqrt a / b ^3 here the 4th root of 81. Well, 1 to the 4th, 2 to the 4th, 3 to the 4th. 1 to the 4th is 1/2 to the fourth is 16. Three to the fourth is 81. If we didn't know that, we could do prime factoring. 81 is 9 * 9, nine is 3 * 3, three times 3. So we'd have the 4th root of 3 to the 4th, X to the fourth over, Y to the 8th Z to the 4th. Four goes into four, one full time, four goes into four again, one full time, Four goes into 8 twice, and four goes into four once. So our final answer is three X / y ^2 Z. This next 1-6 goes into 91 time with some leftovers. 3 leftovers, Six goes into 12 twice with no leftovers, Six goes into 13 twice with one leftover. So AB squared over C ^2 times the 6th root of a ^3 / C divide, and if possible, simplify. We're going to assume all variables represent positive numbers, so here we're going to rewrite it as one single fraction and reduce it. So seven goes into 28 four times and y / y cancels, so sqrt 4 is just two. Here once again, we're going to write it as a single fraction. 56 AB cubed over 7A7 goes into 56. Eight times the A's are going to cancel and we get 8B cubed. Now sqrt 8, sqrt 8 we could think of as 8 being 2 * 4 or 2 * 2 * 2. So the understood 2 index here two goes into three one time with one leftover. The leftovers always go underneath. The radical two goes into the next exponent of three one full time, with one leftover SO2B square root of 2B. This next one we're going to have 189 X to the fifth, Y to the 7th over seven X ^2 y ^2 7 goes into 189 20. 7 * X to the fifth over X ^2 is X ^3 y to the 7th over y ^2 is Y to the fifth. The cube root of 27. Think about prime factoring 27 if you don't know it. So 3 * 9 and 3 * 3 * 3. So the cube root of 3 ^3 X ^3 y to the 5th, the cube root of 3 ^3 is 3 cube root of X ^3 is X. The cube root of Y to the fifth three goes into five one time with two leftover. The 5th root 64 eight of the 11th B to the 28th over 2 AB to the -2. So when we reduce this two goes into 6432 * A to the 11th over a gives us A to the 10th because we're going to subtract those exponents B to the 28th over B to the -2 is going to be B to the 30th because once again, we're going to subtract the exponents 32 four times eight 2 * 2 four times 2. So 32 is 2 to the 5th. And then that 5th root and the 5th power are going to cancel. So we're going to get 25 goes into 10 two hold times with no leftovers, five goes into 36 hold times. So our answer two a ^2 b to the 6th here we're going to rationalize. To rationalize means we need to get a perfect square in the denominator or something. We know the square root up. So I'm going to multiply by sqrt 7 / sqrt 7. That's going to give me sqrt 42 in the top and sqrt 49 in the bottom. But we know that sqrt 49 is just 7. So we're going to multiply, if it's a monomial, by whatever is in the radical and the denominator to rationalize this next one, we have a root 2, so we're going to multiply by a root 2. If I do the bottom, I need to do the top. So we're going to get 3 square roots of 10 / 2 square roots of four. But sqrt 4 is really just two. So we're going to have 2 * 2 in the bottom or three square roots of 10 / 4. This next one, it's different, it's a cube root. So how many do we need to take one out? We need to have three of them. I currently have 1/7. I need two more sevens, so 7 ^2. I have b ^2. How many do I need? Got to have the same as that index. So I need 3. So I need one more B so I'm going to multiply by the cube root of 7 ^2 b and if we do the bottom we have to do the top. So when I multiply this, we're going to get the cube root 3 * 7 ^2 A to the 4th B over the cube root of 7 ^3 b ^3, 7 ^2 is 4949 * 347. We don't have a cube. We don't have 3 threes or three sevens. So I know that that can't split down to get a perfect cube cube root of A to the 4th 1A with one leftover. The cube root of B we don't know. The cube root of 7 ^3 is 7. The cube root of b ^3 is B. This next one, the index is 3 again. I've got 2A's in the bottom, so I need one more. I've got 1B in the bottom, so I need two more. So I have cube root of AB squared over cube root of AB squared. So we get the cube root of 5 AB squared over the cube root of a ^3 b ^3. Well, the cubes and the cube roots cancel, so we're going to get the cube root of 5 AB squared over AB. This next one, let's simplify it first. So 32 four times eight 2 * 2 four times two 2 * 2 * 2 * 2 * 2. So here we're going to have two to the 5th a ^2 b. If we simplify it first, the understood index of two goes into 5/2 times with one leftover and two goes into the a ^2 1 full time with none leftover and the B we only had one of. So now we're going to multiply the top and bottom by sqrt 2 B. So on the top, we're going to get sqrt 14 B. On the bottom we're going to get 4A square roots of 2 ^2 b ^2, which is just going to give us 4A times 2B or 8AB on bottom and sqrt 14 B on top. This next one, let's reduce inside top and bottom first. So there's a three that comes out of the top and the bottom, there's an X that'll reduce in the top and the bottom, and there's AY that'll reduce in the top and the bottom. So if I take a three out of 21, I get 7 and the three out of 75 I get 20. Five X ^2 / X is going to give me X and Y / y to the 5th is going to give me Y to the 4th and the bottom. So sqrt 25 is 5, sqrt y to the 4th is y ^2, and sqrt 7 X we just leave in the top. This next one, we want to rationalize the numerator this time instead of the denominator. So we're going to multiply by sqrt 10 / sqrt 10. That's going to give me sqrt 10 ^2 in the top and sqrt 30 X in the bottom. The square root and the square are going to cancel, so we're going to get 10 / sqrt 30 X this next one, Once again, we're going to rationalize the numerator. We have 1/5 and we need to have three of them because of that index of three. So I'm going to multiply by 5 ^2. If I do the top have to do the bottom, so the cube root of 5 ^3 on top over the cube root of 4 * 5 ^2. The cube root of 5 ^3 is five 4 * 5 ^2. I don't have 3 of anything down here so I know that I can't simplify it any further. We get 5 over the cube root of 100. Thank you and have a wonderful day.