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Simplifying radicals(algebra II)

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    (cube root [tex]\sqrt[]{}x^4[/tex] * [tex]\sqrt[]{}x^5[/tex])^-2

    Honestly, I tried and my answer didnt make any sense...The answer is supposed to be 1/(x^23/3)

    How would I go about solving this? Any help would be appreciated :)

    Also how would I solve the cube root of 7 * the cube root of 49?
  2. jcsd
  3. Feb 3, 2009 #2


    Staff: Mentor

    What you wrote is very difficult to decipher. How about rewriting it using exponents rather than radicals? The cube root of a is a^(1/3).

    The cube root of 7 times the cube root of 49 is the cube root of 343, which can be simplified.
  4. Feb 3, 2009 #3
    Think about what you have: 2 cube roots. One is [tex]\sqrt[3]{7}[/tex] and the other is [tex]\sqrt[3]{7^2}[/tex]. What can you do with that?
    Last edited: Feb 3, 2009
  5. Feb 3, 2009 #4
    ok I get that one! you would get cube root of 7^3...so you would get 7. thanks I get that one now!

    and back to the first problem. in exponents I think it would be:

    (X3/4 * X1/5)-2

    So if anyone wants to help with this one...so far I still have the wrong answer, I have no idea what I'm doing wrong!
  6. Feb 3, 2009 #5


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    Laws of Exponents are taught in Algebra 1 and Algebra 2.

    That one did not type set correctly. I'm trying to state
    'a' to the negative m power equals the fraction one over a to the m power. a^(-m)=1/(a^m)

    Last edited: Feb 3, 2009
  7. Feb 3, 2009 #6


    Staff: Mentor

    That doesn't look like what you started with, which I think should look more like this:
    (x4/3 * something)-2

    The first factor inside the parentheses seems to be the cube root of x^4, which is x4/3. The other factor appears to be the square root of x^5. Can you tell us exactly what the problem is that you're working on?
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