To solve the expression

\[
A = \sqrt[3]{\sqrt{5} + 2} - \sqrt[3]{\sqrt{5} - 2},
\]

we'll denote:

\[
x_1 = \sqrt[3]{\sqrt{5} + 2}, \quad x_2 = \sqrt[3]{\sqrt{5} - 2}.
\]

Thus, we can rewrite \(A\) as:

\[
A = x_1 - x_2.
\]

Next, we will calculate \(A^3\):

\[
A^3 = (x_1 - x_2)^3 = x_1^3 - 3x_1^2 x_2 + 3x_1 x_2^2 - x_2^3 = (x_1^3 - x_2^3) - 3x_1 x_2 (x_1 - x_2).
\]

Now we will evaluate \(x_1^3\) and \(x_2^3\):

\[
x_1^3 = \sqrt{5} + 2, \quad x_2^3 = \sqrt{5} - 2.
\]

Thus,

\[
x_1^3 - x_2^3 = (\sqrt{5} + 2) - (\sqrt{5} - 2) = 4.
\]

Next, we find the product \(x_1 x_2\):

\[
x_1 x_2 = \sqrt[3]{(\sqrt{5} + 2)(\sqrt{5} - 2)} = \sqrt[3]{(\sqrt{5})^2 - 2^2} = \sqrt[3]{5 - 4} = \sqrt[3]{1} = 1.
\]

Now we substitute everything back into our equation for \(A^3\):

\[
A^3 = (x_1^3 - x_2^3) - 3x_1 x_2 A = 4 - 3(1)A \Rightarrow A^3 + 3A - 4 = 0.
\]

Next, we will solve the cubic equation \(A^3 + 3A - 4 = 0\).

Using the Rational Root Theorem, we can test possible rational roots, such as \(A = 1\):

\[
1^3 + 3(1) - 4 = 1 + 3 - 4 = 0.
\]

So \(A = 1\) is a root. Therefore, we can factor the polynomial:

\[
A^3 + 3A - 4 = (A - 1)(A^2 + A + 4).
\]

Next, we need to check the quadratic \(A^2 + A + 4\). The discriminant is:

\[
\Delta = 1^2 - 4 \cdot 1 \cdot 4 = 1 - 16 = -15.
\]

Since the discriminant is negative, \(A^2 + A + 4\) has no real roots.

Thus, the only real solution is

\[
A = 1.
\]

So, the final result is:

\[
\boxed{1}.
\]