We can use the identity for the difference of cubes, which states that:

\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]

In this case, we have \(a = x\) and \(b = y\), so:

\[x^3 - y^3 = (x - y)(x^2 + xy + y^2)\]

Given that \(x - y = 3\) and \(xy = 40\), we can substitute these values into the expression:

\[x^3 - y^3 = (3)(x^2 + xy + y^2)\]

Now we need to find \(x^2 + xy + y^2\). We can square the equation \(x - y = 3\) to get:

\[(x - y)^2 = 3^2\]
\[x^2 - 2xy + y^2 = 9\]

Adding \(4xy\) to both sides gives:

\[x^2 + 2xy + y^2 = 9 + 4xy\]

Substituting \(xy = 40\) gives:

\[x^2 + 2xy + y^2 = 9 + 4(40)\]
\[x^2 + 2xy + y^2 = 9 + 160\]
\[x^2 + 2xy + y^2 = 169\]

Therefore, \(x^2 + xy + y^2 = 169\). Substituting this back into the expression for \(x^3 - y^3\) gives:

\[x^3 - y^3 = (3)(169)\]
\[x^3 - y^3 = 507\]

So, \(x^3 - y^3 = 507\).