Since F is apparently a vector field, I assume you mean
[tex]\vec F = \nabla(x^3+y^3+z^3+3xyz)[/tex]
with ∇ = gradient, whereas ∆ is often used to denote the Laplacian, ∆ = ∂²/∂x² + ∂²/∂y² + ∂²/∂z².
Let [tex]f(x,y,z)=x^3+y^3+z^3+3xyz[/tex]. Compute the gradient of f :
[tex]\vec F = \nabla f(x,y,z) = \dfrac{\partial f}{\partial x} \, \vec\imath + \dfrac{\partial f}{\partial y} \, \vec\jmath + \dfrac{\partial f}{\partial z} \, \vec k[/tex]
[tex]\vec F = (3x^2+3yz) \,\vec\imath + (3y^2 + 3xz) \,\vec\jmath + (3z^2+3xy) \,\vec k[/tex]
Now compute the divergence of F (incidentally, divergence of a gradient field is the Laplacian of the f):
[tex]\mathrm{div} \, \vec F = \dfrac{\partial(3x^2+3yz)}{\partial x} + \dfrac{\partial(3y^2+3xz)}{\partial y} + \dfrac{\partial(3z^2+3xy)}{\partial z}[/tex]
[tex]\boxed{\mathrm{div} \, \vec F = 6x + 6y + 6z}[/tex]
and the curl: (the following is overkill, since any gradient field has curl zero, but it doesn't hurt to verify that)
[tex]\mathrm{curl}\, \vec F = \left(\dfrac{\partial(3z^2+3xy)}{\partial y} - \dfrac{\partial(3y^2+3xz)}{\partial z}\right) \,\vec\imath - \left(\dfrac{\partial(3z^2+3xy)}{\partial x} - \dfrac{\partial(3x^2+3yz)}{\partial z}\right) \, \vec\jmath \\ ~~~~~~~~~~~~ + \left(\dfrac{\partial(3y^2+3xz)}{\partial x} - \dfrac{\partial(3x^2+3yz)}{\partial y}\right) \,\vec k[/tex]
[tex]\boxed{\mathrm{curl} \,\vec F = \vec 0}[/tex]
