Given a vector function $\mathbf{R} = x\mathbf{i}+ y\mathbf{j}+z\mathbf{k}$ and $r = |\mathbf{R}| = \sqrt{x^2 + y^2 + z^2}$ then prove that for any vector constant $\mathbf{A}$ holds $\nabla\times\Big(\frac{1}{r}(\mathbf{A}\times\mathbf{R})\Big)=\frac{1}{r}\mathbf{A}+\frac{\mathbf{A}\cdot\mathbf{R}}{r^2}\mathbf{R}$
This is my friend answer but i dont understand Is he right?
$\nabla\times\Big(\frac{1}{r}(\mathbf{A}\times\mathbf{R})\Big)$ \begin{align*} ={}&\Big(\frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k\Big)\times\Big(\frac{1}{r}(A_2z-A_3y)i+(A_3x-A_1z)j\\&+(A_1y-A_2x)k\Big)\\ ={}& \left| \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{1}{r}(A_2z-A_3y) & \frac{1}{r}(A_3x-A_1z) & \frac{1}{r}(A_1y-A_2x) \end{array} \right|\\ ={}& \Big(\frac{\partial}{\partial y}\Big(\frac{1}{r}(A_1y-A_2x)\Big)-\frac{\partial}{\partial z}\Big(\frac{1}{r}(A_2z-A_3y)\Big)\Big)i + \\&\Big(\frac{\partial}{\partial z}\Big(\frac{1}{r}(A_2z-A3_y)\Big)-\frac{\partial}{\partial x}\Big(\frac{1}{r}(A_1y-A_2x)\Big)\Big)j\\& + \Big(\frac{\partial}{\partial x}\Big(\frac{1}{r}(A_3x-A_1z)\Big)-\frac{\partial}{\partial y}\Big(\frac{1}{r}(A_2z-A_3y)\Big)\Big)k\\ ={}&\Big(\frac{\partial}{\partial y}\Big(\frac{1}{\sqrt{x^2 + y^2+z^2}}(A_1y-A_2x)\Big)-\frac{\partial}{\partial z}\Big(\frac{1}{\sqrt{x^2 + y^2+z^2}}(A_2z-A_3y)\Big)\Big)i\\& +\Big(\frac{\partial}{\partial z}\Big(\frac{1}{\sqrt{x^2 + y^2+z^2}}(A_2z-A_3y)\Big)-\frac{\partial}{\partial x}\Big(\frac{1}{\sqrt{x^2 + y^2+z^2}}(A_1y-A_2x)\Big)\Big)j\\& + \Big(\frac{\partial}{\partial x}\Big(\frac{1}{\sqrt{x^2 + y^2+z^2}}(A_3x-A_1z)\Big)-\frac{\partial}{\partial y}\Big(\frac{1}{\sqrt{x^2 + y^2+z^2}}(A_2z-A_3y)\Big)\Big)k\\ ={}& \Big(\Big((A_1y-A_2x)\frac{\partial}{\partial y}\Big(\frac{1}{\sqrt{x^2 +y^2+z^2}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)\frac{\partial}{\partial y}(A_1y-A_2x)\Big)-\\& \Big(\Big((A_3x-A_1z)\frac{\partial}{\partial z}\Big(\frac{1}{\sqrt{x^2 +y^2+z^2}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)\frac{\partial}{\partial z}(A_3x-A_1z)\Big)\Big)i\\& +\Big(\Big((A_2z-A_3y)\frac{\partial}{\partial z}\Big(\frac{1}{\sqrt{x^2 +y^2+z^2}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)\frac{\partial}{\partial z}(A_2z-A_3y)\Big)-\\& \Big(\Big((A_1y-A_2x)\frac{\partial}{\partial x}\Big(\frac{1}{\sqrt{x^2 +y^2+z^2}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)\frac{\partial}{\partial x}(A_1y-A_2x)\Big)\Big)j+\\& \Big(\Big((A_3x-A_1z)\frac{\partial}{\partial z}\Big(\frac{1}{\sqrt{x^2 +y^2+z^2}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)\frac{\partial}{\partial z}(A_3x-A_1z)\Big)-\\& \Big(\Big((A_2z-A_3y)\frac{\partial}{\partial y}\Big(\frac{1}{\sqrt{x^2 +y^2+z^2}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)\frac{\partial}{\partial y}(A_2z-A_3y)\Big)\Big)k\\ ={}& \Big(\Big((A_1y-A_2x)\Big(\frac{-y}{(x^2 +y^2+z^2)^{\frac{3}{2}}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)A_1\Big)+\\& \Big(\Big((A_3x-A_1z)\Big(\frac{z}{(x^2 +y^2+z^2)^{\frac{3}{2}}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)A_1\Big)\Big)i\\& +\Big(\Big((A_2z-A_3y)\Big(\frac{-z}{(x^2 +y^2+z^2)^{\frac{3}{2}}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)A_2\Big)+\\& \Big(\Big((A_1y-A_2x)\Big(\frac{x}{(x^2 +y^2+z^2)^{\frac{3}{2}}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)A_2\Big)\Big)j\\& +\Big(\Big((A_3x-A_1z)\Big(\frac{-x}{(x^2 +y^2+z^2)^{\frac{3}{2}}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)A_3\Big)+\\& \Big(\Big((A_2z-A_3y)\Big(\frac{y}{(x^2 +y^2+z^2)^{\frac{3}{2}}}\Big)+\Big(\frac{1}{\sqrt{x^2+y^2+z^2}}\Big)A_3\Big)\Big)k\\ ={}& \frac{A_1}{\sqrt{x^2+y^2+z^2}}i+\frac{A_2}{\sqrt{x^2+y^2+z^2}}j+\frac{A_3}{\sqrt{x^2+y^2+z^2}}k\\& +\frac{\Big((-A_1y^2+A_2xy)+(A_3xz-A_1xz^2)\Big)}{(x^2+y^2+z^2)^\frac{3}{2}}i+\frac{A_1}{\sqrt{x^2+y^2+z^2}}i\\& +\frac{\Big((-A_2z^2+A_3yz)+(A_1xy-A_2x^2)\Big)}{(x^2+y^2+z^2)^\frac{3}{2}}j+\frac{A_2}{\sqrt{x^2+y^2+z^2}}j\\& +\frac{\Big((-A_3x^2+A_1xz)+(A_2yz-A_3y^2)\Big)}{(x^2+y^2+z^2)^\frac{3}{2}}k+\frac{A_3}{\sqrt{x^2+y^2+z^2}}k\\ ={}&\frac{A_1}{\sqrt{x^2+y^2+z^2}}i+\frac{A_2}{\sqrt{x^2+y^2+z^2}}j+\frac{A_3}{\sqrt{x^2+y^2+z^2}}k\\& +\frac{A_1x+A_2y+A_3z}{\sqrt{x^2+z^2+z^2}^\frac{4}{2}}(xi+yj+zk)\\ ={}&\frac{A_1i+A_2j+A_3k}{\sqrt{x^2+y^2+z^2}}+\frac{A_1x+A_2y+A_3z}{\Big(\sqrt{x^2+y^2+z^2}\Big)^2}(xi+yj+zk)\\ ={}&\frac{1}{\sqrt{x^2+y^2+z^2}}\Big(A_1i+A_2j+A_3k\Big)+\frac{(A_1i+A_2j+A_3k)(xi+yj+zk)}{\Big(\sqrt{x^2+y^2+z^2}\Big)^2}(xi+yj+zk)\\ ={}&\frac{1}{r}A+\frac{AR}{r^2}R \end{align*}