Since F is apparently a vector field, I assume you mean
\(\vec F = \triangledown(x^3 + y^3 + z^3+ 3xyz)\)
With ∇ = gradient, whereas ∆ is often used to denote the Laplacian, \(\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial ^2}{\partial y^2} + \frac{\partial ^2}{\partial z^2}\).
Let \(f(x, y, z) = x^3 + y^3 + z^3 + 3xyz\).
Compute the gradient of f:
\(\vec F = \Delta f(x, y,z) = \frac{\partial f}{\partial x}\vec i + \frac{\partial f}{\partial y}\vec j+ \frac{\partial f}{\partial z}\vec k\)
\(\vec F = (3x^2 + 3yz)\vec i + (3y^2 + 3xz) \vec j + (3z^2 + 3xy) \vec k\)
Now compute the divergence of F (incidentally, divergence of a gradient field is the Laplacian of the f):
\(\text{div} \,\vec F = \frac{\partial (3x^2 +3yz)}{\partial x} + \frac{\partial (3y^2 +3xz)}{\partial y} + \frac{\partial (3z^2 +3xy)}{\partial z}\)
\(\text{div} \,\vec F =6x + 6y + 6z\)
And the curl:
\(\text{curl } \vec F= \left(\frac{\partial (3z^2 + 3xy)}{\partial y} - \frac{\partial (3y^2 + 3xz)}{\partial z} \right) \vec i - \left(\frac{\partial (3z^2 + 3xy)}{\partial x} - \frac{\partial (3x^2 + 3yz)}{\partial z} \right) \vec j +\left(\frac{\partial (3y^2 + 3xz)}{\partial x} - \frac{\partial (3x^2 + 3yy)}{\partial z} \right) \vec k \)
\(\text{curl } \vec F= \vec0\)
