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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 $\displaystyle (x^2+y^2+z^2)^{3/2}+3x^2(x^2+y^2+z^2)^{5/2}=\frac{2x^2y^2z^2}{(x^2+y^2+z^2)^{5/2}}$ James Last edited by bobred; February 2nd, 2015 at 06:23 AM. 
February 2nd, 2015, 07:19 AM  #2  
Member Joined: Jan 2015 From: Rajasthan, India Posts: 55 Thanks: 24 Math Focus: Algebra & Geomectry  Quote:
$\displaystyle ((x^2+y^2+z^2)^{1/2})^{3}+3x^2((x^2+y^2+z^2)^{1/2})^{5}$ 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 $\displaystyle \frac{1}{K^5}\dot{{3x^2}{{x^2}+{y^2}+{z^2}}}$ = $\displaystyle \frac{1}{K^5}\dot{{2x^2}{y^2}{z^2}}$ =$\displaystyle \frac{2x^2y^2z^2}{(x^2+y^2+z^2)^{5/2}}$ = RHS  
February 2nd, 2015, 07:35 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,805 Thanks: 2150 
$\displaystyle (x^2+y^2+z^2)^{3/2}+3x^2(x^2+y^2+z^2)^{5/2}$ $\displaystyle =\frac{(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{5/2}} + \frac{3x^2}{(x^2+y^2+z^2)^{5/2}}$ $\displaystyle = \frac{2x^2y^2z^2}{(x^2+y^2+z^2)^{5/2}}$ 

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