Q.4 (c) If \(u=\log(x^3+y^3+z^3-3xyz)\), prove that \(\left(\frac\partial{\partial x}+\frac\partial{\partial y}+\frac\partial{\partial z}\right)^2u=-\frac9{(x+y+z)^2}\).
Best answer
\(u_x=\frac{3x^2-3yz}{x^3+y^3+z^3-3xyz},\ u_y=\frac{3y^2-3zx}{\cdots},\ u_z=\frac{3z^2-3xy}{\cdots}\)
\(u_x+u_y+u_z=\frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}=\frac3{x+y+z}\)
(using \(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\)).
\(\left(\frac\partial{\partial x}+\frac\partial{\partial y}+\frac\partial{\partial z}\right)^2u=\left(\frac\partial{\partial x}+\frac\partial{\partial y}+\frac\partial{\partial z}\right)\frac3{x+y+z}=-\frac3{(x+y+z)^2}\times3\)
\(=-\frac9{(x+y+z)^2}\) (proved).
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