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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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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 $\displaystyle -(x^2+y^2+z^2)^{-3/2}+3x^2(x^2+y^2+z^2)^{-5/2}$ $\displaystyle =\frac{-(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{5/2}} + \frac{3x^2}{(x^2+y^2+z^2)^{5/2}}$ $\displaystyle = \frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$

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