1
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Laplacian[x/(x^2 + y^2), {x, y}]
(8 x^3)/(x^2 + y^2)^3 + (8 x y^2)/(x^2 + y^2)^3 - (8 x)/(x^2 + y^2)^2
% === 0

False

Since we have $$\frac{8x^3}{(x^2+y^2)^3}+\frac{8xy^2}{(x^2+y^2)^3}-\frac{8x}{(x^2+y^2)^2}$$

we can turn the third term to $\frac{(x^2+y^2)8x}{(x^2+y^2)(x^2+y^2)^2}$, and simplify this whole expression to get zero.

How can I get Mathematica to do this?

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closed as off-topic by Bob Hanlon, m_goldberg, J. M. will be back soon Mar 19 '17 at 5:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – Michael E2 Mar 19 '17 at 3:07
  • $\begingroup$ People here generally like users to post code as Mathematica code instead of images or TeX -- and certainly not as W|A url input -- so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful $\endgroup$ – Michael E2 Mar 19 '17 at 3:08
  • $\begingroup$ @MichaelE2 Sorry Michael, I'm a newbie in Mathematica.SE! I have updated my question. $\endgroup$ – Chenyao Lou Mar 19 '17 at 3:21
  • $\begingroup$ You have to simplify the expression by using Simplify command. $\endgroup$ – Anjan Kumar Mar 19 '17 at 3:24
  • 2
    $\begingroup$ Checking for equality doesn't simplify expressions to the full. For example, x-x===0 returns True, whereas, 1/(1 + a) - a/(a + a^2) === 0 returns False. Therefore, it is better to explicitly use Simplify. $\endgroup$ – Anjan Kumar Mar 19 '17 at 3:37