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Math Help - cuberooting

  1. #1
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    cuberooting

    I have a few questions n stuff thats holding me back in math because I don't understand why the problems are worked this way I have the cuberoot of 7 over the cuberoot of 12. 3SQRTSIGN 7/3SQRTSIGN 12
    now how do I rationalize the denominator. The answer is cuberoot of 126 / 6. How can I do problems like these without a calculator?
    Also assuming all variables are positive how do I simplify these two problems
    [can someone write clear step by step ways on how you did this?] i know i'm supposed to multiply everything inside the parenthesis by the number outside right?
    (x^-4/7)^7 and (3x^2/3)^-1 THANKS
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by D@nny View Post
    I have a few questions n stuff thats holding me back in math because I don't understand why the problems are worked this way I have the cuberoot of 7 over the cuberoot of 12. 3SQRTSIGN 7/3SQRTSIGN 12
    now how do I rationalize the denominator. The answer is cuberoot of 126 / 6. How can I do problems like these without a calculator?
    Also assuming all variables are positive how do I simplify these two problems
    [can someone write clear step by step ways on how you did this?] i know i'm supposed to multiply everything inside the parenthesis by the number outside right?
    (x^-4/7)^7 and (3x^2/3)^-1 THANKS
    note that our objective when rationalizing the denominator is to get a number in the bottom with a power of 1. so we multiply by the denominator (to an appropriate power) over itself to accomplish this

    here's the problem:

    keeping every thing in radical signs.

    cuberoot(7) / cuberoot(12)
    = cuberoot(7) / cuberoot(12) * (cuberoot(12))^2 / (cuberoot(12))^2
    = cuberoot(7)(cuberoot(12))^2 / (cuberoot(12))^3
    = cuberoot(7)(cuberoot(12^2)) /12 ..........note that i moved the square inside the cuberoot, why can we do this? you'll see it more explicitly in the next method i use

    = cuberoot(7*(12^2))/12 ........i combined the cuberoots using a law of surds (and exponents)
    = cuberoot(1008)/12
    = cuberoot(8*126)/12
    = cuberoot(8)*cuberoot(126)/12
    = 2*cuberoot(126)/12
    = cuberoot(126)/6

    second method, changing radical signs to powers:

    cuberoot(7) / cuberoot(12)
    = 7^(1/3) / 12^(1/3)
    = 7^(1/3) / 12^(1/3) * 12^(2/3) / 12^(2/3) .......i multiply by 12^2/3 since i know when i multiply 12^(1/3) by 12^(2/3) i will add the powers. 1/3 + 2/3 = 1, which is what we want

    = [7^(1/3) * 12^(2/3)] / [12^(1/3) * 12^(2/3)]
    = [7^(1/3) * 12^{2*(1/3)}] / [12^(1/3) * 12^(2/3)] .....now you see why i could move the square in and out, because as powers, we have the multiple 2/3. but 2/3 = 2*(1/3) = (1/3)*2. the 2 gives the square, the 1/3 gives the cuberoot. this means i have the choice of taking the cuberoot first and the square later, or the square first and the cuberoot later
    = [7^(1/3) * (12^2)^(1/3)}] / [12^(1/3 + 2/3)]
    = [{7*(12^2)}^(1/3)] / [12^(1/3) * 12^(2/3)] .......this is a law of exponents... (x^m)*(y^m) = (xy)^m
    = 1008^(1/3) / 12
    = (126*8)^(1/3) / 12
    = [8^(1/3) * 126^(1/3)]/12
    = 2*126^(1/3) / 12
    = 126^(1/3) / 12
    = cuberoot(126) / 12
    Last edited by Jhevon; April 12th 2007 at 06:17 PM.
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