Solution:
Cubing both sides of the given equation yields
$$x + 3 \sqrt[3]{x(4-x)} (\sqrt[3]{x} + \sqrt[3]{4-x}) + 4 - x = 1,$$
which then becomes
$$4 + 3 \sqrt[3]{x(4-x)} = 1$$
or
$$\sqrt[3]{x(4-x)} = -1.$$
Cubing both sides of this equation gives
$$x(4-x) = -1$$
or
$$x^2 - 4x = 1.$$
This means
$$(x-2)^2 = 5$$
so
$$x = 2 \pm \sqrt{5}$$
and the largest real solution is $x = 2 + \sqrt{5}$.