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February 2nd, 2015, 06:16 AM   #1
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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.
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February 2nd, 2015, 07:19 AM   #2
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Quote:
Originally Posted by bobred View Post
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
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February 2nd, 2015, 07:35 AM   #3
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$\displaystyle -(x^2+y^2+z^2)^{-3/2}+3x^2(x^2+y^2+z^2)^{-5/2}$

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

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