how to show that $\Delta [\frac{c- |x-a|^2}{|x-b|^n}] = 2n|x-b|^{-n-2}(c-|a-b|^2), \; x \neq b$

Question

$a, b, x \in \mathbb{R^n}, \; c \in \mathbb{R}$

$$\nabla(\cdot) = [{ \partial \over \partial x_1}(\cdot)\; , \;{ \partial \over \partial x_2} (\cdot) \;,\;\dots \; , \; { \partial \over \partial x_n} (\cdot)]$$

$$\Delta(\cdot) = [{ \partial \over \partial x_1} \circ {\partial \over \partial x_1} + { \partial \over \partial x_2} \circ {\partial \over \partial x_2} + \dots+{ \partial \over \partial x_n} \circ {\partial \over \partial x_n}](\cdot)$$

I've already established two potentially useful equalities which are :

$$\nabla [|x-a|^\alpha ]= \alpha|x-a|^{\alpha-2}(x-a)$$

$$\Delta[|x-a|^{\alpha}] = \alpha(n+\alpha-2)|x-a|^{\alpha -2}$$

using the fact that $\Delta[fg] = f\Delta g + 2\nabla f \cdot\nabla g +g\Delta f$

farthest I could reach was :

$$\Delta [\frac{c- |x-a|^2}{|x-b|^n}] = 2nc|x-b|^{-n-2} - \Delta[\frac{|x-a|^2}{|x-b|^n}]$$

and

$$\Delta [\frac{|x-a|^2}{|x-b|^n}] = 2n|x-b|^{-n} + 2\nabla f \cdot \nabla g + 2n|x-b|^{-n-2}|x-a|^2$$

I'm having difficulties tackling the product of gradients.

any comments, hints will be greatly appreciated.

Answer 1

In fact I should've just used the fact that : $|v - w|^2= |v|^2+|w|^2 -2v\cdot w$

with $v = x-a, \; w = x-b$

let's do it : $$\begin{align} & 2n|x-b|^{-n} + 2\nabla f \cdot \nabla g + 2n|x-b|^{-n-2}|x-a|^2 \\ =& 2n|x-b|^{-n-2}[|x-b|^2+|x-a|^2] + 2\nabla f \cdot \nabla g \\ =& 2n|x-b|^{-n-2}[|x-b|^2+|x-a|^2] + 2[2(-n)|x-b|^{-n-2}(x-a)\cdot(x-b)] \\ =& 2n|x-b|^{-n-2}[|x-b|^2+|x-a|^2 - 2] (x-a)\cdot(x-b)] \\ =& 2n|x-b|^{-n-2}|a-b|^2 . \end{align}$$

Answer 2

Hint: Noting $$ \nabla|x-a|^2=2(x-a), \nabla|x-b|^{-n}=-n|x-b|^{-n-2}(x-b) $$ one has $$\nabla|x-a|^2\nabla|x-b|^{-n}=-2n|x-b|^{-n-2}(x-a)(x-b)$$