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 February 2nd, 2015, 06:16 AM #1 Newbie   Joined: Jun 2009 Posts: 5 Thanks: 0 Simplifying an equation Hi I am not sure where to start when trying to simplify the LHS so it equals the RHS $\displaystyle -(x^2+y^2+z^2)^{-3/2}+3x^2(x^2+y^2+z^2)^{-5/2}=\frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ James Last edited by bobred; February 2nd, 2015 at 06:23 AM. February 2nd, 2015, 07:19 AM   #2
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Math Focus: Algebra & Geomectry Quote:
 Originally Posted by bobred Hi I am not sure where to start when trying to simplify the LHS so it equals the RHS $\displaystyle -(x^2+y^2+z^2)^{-3/2}+3x^2(x^2+y^2+z^2)^{-5/2}=\frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ James
you can write left part of equation as
$\displaystyle -((x^2+y^2+z^2)^{1/2})^{-3}+3x^2((x^2+y^2+z^2)^{-1/2})^{-5}$

Now, for simplicity, assume $\displaystyle (x^2+y^2+z^2)^{1/2}$ = K
Therefore, eq. become
$\displaystyle \frac{3x^2}{K^5}-\frac{1}{K^3}$
Now, take common $\displaystyle \frac{1}{K^5}$ from both terms-
we get, $\displaystyle \frac{1}{K^5}\dot{{3x^2}-{K^2}}$

Since, K$\displaystyle ^2$ = $\displaystyle {x^2}+{y^2}+{z^2}$

therefore, we get $\displaystyle \frac{1}{K^5}\dot{{3x^2}-{{x^2}+{y^2}+{z^2}}}$

= $\displaystyle \frac{1}{K^5}\dot{{2x^2}-{y^2}-{z^2}}$

=$\displaystyle \frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$

= RHS February 2nd, 2015, 07:35 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2272 $\displaystyle -(x^2+y^2+z^2)^{-3/2}+3x^2(x^2+y^2+z^2)^{-5/2}$ $\displaystyle =\frac{-(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{5/2}} + \frac{3x^2}{(x^2+y^2+z^2)^{5/2}}$ $\displaystyle = \frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ Tags equation, simplifying Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Kinh Algebra 5 March 13th, 2014 11:00 PM bytelogik Algebra 3 June 17th, 2012 06:01 PM jakeward123 Algebra 4 May 11th, 2012 12:37 PM cuteascanb Algebra 4 January 9th, 2011 05:45 AM cuteascanb New Users 1 December 31st, 1969 04:00 PM

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