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 March 6th, 2014, 06:06 PM #1 Newbie   Joined: Mar 2014 Posts: 1 Thanks: 0 cube root What is the cube root of (81x^8y^12)? It is probably super easy, but after several tries, I just give up. Thank you so much!!! Given answers: A) 3x^2y^4 cube root of (3x^2) B) 3x^2y^4 fifth root of (3x^2) C) 3x^2y^12 cube root of (3x^2) D) 3x^2y^4 cube root of (3xy)
 March 6th, 2014, 06:29 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra Re: cube root The cube root of any product is the product of the cube root of each elements of the product. That is: $\sqrt[3]{abcd}=\sqrt[3]{a}\sqrt[3]{b}\sqrt[3]{c}\sqrt[3]{d}$ Also $\sqrt[3]{x}= x^{\frac{1}{3}}$ and $(x^{a})^b= x^{ab}$ Armed with that, you should be able to have another go.
 March 6th, 2014, 06:40 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: cube root $81= 9(9)= 3(3)(3)(3)= 3^3(3)$ $x^8= x^{6+ 2}= x^6x^2= (x^2)^3x^2$ $y^{12}= x^{3(4)}= (x^4)^3$
 March 6th, 2014, 07:13 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: cube root $(81x^8y^{12})^{\frac13}\,=\,(3^4x^8y^{12})^{\frac1 3}\,=\,3^{\frac43}\,\cdot\,x^{\frac83}\,\cdot\,y^{ \frac{12}{3}}\,=\,3^{\frac13}\,\cdot\,3^1\,\cdot\, x^2\,\cdot\,x^{\frac23}\,\cdot\,y^4\,=\,3x^2y^4(3x ^2)^{\frac13}$ When multiplying exponentials with the same base, add the exponents. The above uses that rule "in reverse".
March 6th, 2014, 11:57 PM   #5
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Quote:
 Originally Posted by lila342 . . . after several tries, I just give up.
There are only four choices, and the second choice is absurd (the third choice also, if given a moment's thought), so how could you need more than three tries?

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