Problem statement: If $\displaystyle u=\frac{1}{r}$ where $\displaystyle r=\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}$, for $\displaystyle r\ne0$ prove that $\displaystyle \bigtriangleup u \equiv \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=0$

This is my approach: $\displaystyle \frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{dr}{dx}=-\frac{x-a}{r^2\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}$

$\displaystyle \frac{\partial^2 u}{\partial x^2}=\frac{\partial(\frac{\partial u}{\partial x})}{\partial x}=\frac{\partial(\frac{\partial u}{\partial x})}{\partial r}\frac{dr}{dx}=\frac{2(x-a)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}$

$\displaystyle \frac{\partial^2 u}{\partial y^2}=\frac{\partial(\frac{\partial u}{\partial y})}{\partial y}=\frac{\partial(\frac{\partial u}{\partial y})}{\partial r}\frac{dr}{dy}=\frac{2(y-b)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}$

$\displaystyle \frac{\partial^2 u}{\partial z^2}=\frac{\partial(\frac{\partial u}{\partial z})}{\partial z}=\frac{\partial(\frac{\partial u}{\partial z})}{\partial r}\frac{dr}{dz}=\frac{2(z-c)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}$

So $\displaystyle \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=\frac{2(x-a)^2+2(y-b)^2+2(z-c)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}=\frac{2}{r}\ne0$

It seems I made an major error somewhere or I have conceptual problems.

Help me!

Thank you