Split the quotient: 27^(1/3) * (1/16)^(1/3). 16 can be written as: 2 * 8. That means: 3 * (1/(2 * 8))^(1/3). 2^3 = 8. That means that the cube root of 8 is 2. So, the equation can also be written like: 3 / (2 * 2^(1/3)). Or: (3/2) / (2^(1/3)). Or: 1.5 * 2^(-1/3).
You all can look at it this way. When you have an a^(1/r) in the denominator, you need to give this term r-1 identical "buddies" and then put a copy of each of those "buddies" in the numerator to work toward the proper radical format when presenting a value as an answer. If it is a square root, you need one (2 - 1) "a^(1/2)" in both the numerator and denominator. if it were a seventh root then you, of course, would need six (7 - 1) "a^(1/7)"s in both the numerator and denominator.
I would have first simplified it as: 3*cube_root(1/16) using the multiplication property of radicals, reading 27/16 as 27 * 1/16. Then you can reduce cube_root(1/16) by using the equivalent fraction 4/64, and the division property of radicals to get 3*cube_root(4)/4
I didn't know how to rationalize a cube root denominator before watch your solution I was looking for ways to solve it this is what I thought the cube root of 16 is 16^1/3 so if I times both numerator and denominator by 16^2/3 it would leave 16^3/3 so just 16 3x 16^2/3 = 3 x cube root 256 3 x cube root of 64x4 3x4x cube of 4 / 16 3*Cube root 4 / 4 Your way is much easier
It would make more sense to, in the last step, multiply by the cube root of 2 over the cube root of 2 twice...effectively cubing the cube root of 2 in the denominator to give 2; then simplifying the 2 cube roots of 2 in the numerator to the cube root of 4.
The way I've seen it in classes and on these math channels, is they tend to frown on having fractional exponents in the final answer, or exponents in the radicand, unless logistically necessary (ie: if it's more feasible to express the radicand as an exponentiated term rather than a rational number). Thus ∛4 is more ideal than 2^(2/3) or ∛2², but ∜71³ would be preferable to ∜357911.
I got part way through doing this in my head and saw your cube root of 4 in the answer and said, "What the ...???" I was thinking of just multiplying the cube root of 2 times the cube root of 2. But I was wrong, as you showed. Doing that will still leave a radical in the denominator. Oops! Good problem and good explanation. But please try not to repeat yourself so much. You explained the 7/√3 example three times.
Well explained! I got to 3/the cube root of 16 but I knew it was not enough. Repeat after me, we can't have an irrational number as the denominator. :( Got it! Thanks!
My math OCD doesn't like cubes in the numerator. I'd still like to solve it, as it seems like it doesn't have a "true" final answer. It would have a decimal that never ends, most likely. 🤔
Super simple I will use fraction exponents to make it easier to read. Numerator first: 27 ^ 1/3 = (3 * 3 * 3) ^ 1/3 = 3 This shows that 27 is a perfect Cube Denominator: 16 ^ 1/3 = (2 * 2 * 2 * 2) ^ 1/3 = 2(2^1/3) This shows 16 is NOT a perfect cube but since 8 is we can rewrite 16^1/3 as (8 * 2) ^ 1/3. Which gives me 2(2^ 1/3) 2(2 * 4) ^ 1/3 = 2(8 ^ 1/3) = 2 * 2 or just 4 We have a root in the denominator which is a no no. So we have to think of the smallest number that when multiplied by 2 will give me a perfect cube. We know that 2^3 = 8 so we can turn 2^1/3 into 8 ^ 1/3 by multiplying which gives us (2 * 4) ^ 1/3 Numerator: 3(4 ^ 1/3) Because I multiplied the denominator by x I also have to multiply the numerator by x Final Answer: [3(4^1/3)] / 4 cube root 4 can NOT cross cancel with the denominator of 4 so that is fully simplified.
I didn't get it right at first, but I understand your explanation. I am a 71-year-old senior citizen student at my local community college, with a 3.9 GPA. 😊
@@jamesharmon4994 Why? I have run into many equations used in engineering and science that have radicals in the denominator. If it's good enough to design bridges and spacecraft, it's good enough for any Real World application.
Disagree. Acceptable answer is 3/(2*cube root(2)). Reason, Sin 45 is 1/sqrt(2) not sqrt(2)/2. It is acceptable to have radicals in the denominator, well, it was when I was at secondary school 55 years ago.
I'm getting (3*³√2)/4 oops should have multiplied ³√4/³√4 for my "one" used ³√2/³√2 instead (thinking squares instead of cubes. Correct answer is (3*³√4)/4.😂
For this comment, I’ll not even try to solve the question. Just a thought. ‘Many will get wrong’. In all vids. True statement, but it is phrased in the negative, missing an ‘it’ I guess on purpose, some psychology coming in… Personally, not a creator on RU-vid , I’d go with ‘Did you get it right?’. Feels much more engaging to me. Just my thoughts, let the shredding begin, as long as it stays within the boundaries of math.
If your're trying to write 3/2*cuberoot(2), thats correct but you have to rationalize the denominator. To do that we multiply top and bottom by cuberoot(2)^2. cuberoots must be multiplied out 3 times unlike square roots. So apply that to the denominator, you get 2cuberoot(2) * cuberoot(2)^3 which gives 4 and on top multiply 3*cuberoot(2)^2, we square it here because there was not a cuberoot(2) already there like in the denominator. this gives 3cuberoot(4)/2
its (3*cuberoot(4))/4. What you have is correct but you have to rationalize the denominator. Since we are dealing with cuberoots we have to multiply top and bottom by cuberoot(2) twice (cuberoot(2)*cuberoot(2)*cuberoot(2))=2 and on top you have 3*(cuberoot(2))^2 = (3cuberoot(4)/4)
Actually 1 1/2 * ((1/2)^1/3) is not an invalid answer. Just because this guy chose to leave the radical in the denominator doesn't mean that it had to be removed in other solutions to the problem.
@@Shay-q8u Well, the task was explicitly: "simplify". 60 years ago 3*CubeRoot(4)/4 was certainly simpler to calculate than 3/(2*CubeRoot(2)), using a logarithm table -- but today, using a spreadsheet?
I really enjoyed math throughout K-12 and into my graduate studies. I truly enjoy the reviews and the mental exercises. However, your constant talking about things that do not apply to the problem at hand runs my blood pressure up to the point that I cannot listen or watch.
But if you present your short solution to a student, it will not teach them how to arrive at your solution. John is teaching, not just showing the way an advanced level mathematician might do it in his head.
It's a great question. Unfortunately, most teachers never explain why we teach certain things. The answer, BTW, to your question is a resounding 'Never'. However, this is not the point. The reason this is taught is because every time your brain is confronted with a challenge it must create new neurological connections in order to find a path to the answer. After a few years of schooling, if you allow this process to take place you will (hopefully) end up with a brain that is better able to seek out solutions.
@@mathmandrsam Well, that might be. That said, in my professional life, which includes 36 years of tax practice, and 26 years of teaching law, well, cube roots have never come up in any conversation.
@@louf7178 Yup!!! Basket Weaving, Tax, Business Law, Finance, Management, Investment Analysis, to name just a few. None of that stuff is relevant to those subject areas, and quite a few more, I would guess.
