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May 15th, 2008, 01:21 AM
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| |  radical expressions/exponents.
mmkay so im new at this..but i cant figure these out to save my life.. and i have a test in about six hours which is what either make me pass or fail the class..so herei go:
simplify:
the cube root of 'A' to the fourth over nine. |
May 15th, 2008, 02:10 AM
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Quote:
Originally Posted by morgan.  mmkay so im new at this..but i cant figure these out to save my life.. and i have a test in about six hours which is what either make me pass or fail the class..so herei go:
simplify:
the cube root of 'A' to the fourth over nine. | You mean ![(\sqrt[3]{A})^{\frac49}](http://www.mathhelpforum.com/math-help/latex2/img/85da6f995b0363afe369e893d8bfe542-1.gif "(\sqrt[3]{A})^{\frac49}") ? ![(\sqrt[3]{A})^{\frac49} = ((A)^{\frac13})^{\frac49} = (A)^{{\frac13}{\frac49}} = A^{\frac4{27}}](http://www.mathhelpforum.com/math-help/latex2/img/446e84a9cd8b33066a2f32758b3cebc1-1.gif "(\sqrt[3]{A})^{\frac49} = ((A)^{\frac13})^{\frac49} = (A)^{{\frac13}{\frac49}} = A^{\frac4{27}}")
But I doubt this is a simplification... looks like a complication.... 
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May 15th, 2008, 06:18 AM
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How do we know it's not: ![\frac{(\sqrt[3]a)^4}{9}](http://www.mathhelpforum.com/math-help/latex2/img/0fb050494a935fe17070d79dc0e8e169-1.gif "\frac{(\sqrt[3]a)^4}{9}")
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