VC3.31

$\nabla^2 u=[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}]u\,$

= $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}\,$ ---(1)

Now $\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}(ax^2+by^2+cz^2)=2ax\,$

$\frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}(2ax)=2a\,$

Similarly, $\frac{\partial^2 u}{\partial y^2}=2b,\frac{\partial^2 u}{\partial z^2}=2c\,$

So,(1)= $\nabla^2 u=2a+2b+2c=2(a+b+c)\,$

Hence $\nabla^2 u=0\,$ provided $a+b+c=0\,$