If the area of the region $$\left\{(x,y) : x^{\frac{2}{3}} + y^{\frac{2}{3}} \leq 1, x + y \geq 0, y \geq 0\right\}$$ is $$A$$, then $$\frac{256A}{\pi}$$ is ______

Given region,

$$x^{\frac23}+y^{\frac23}\le1,\qquad x+y\ge0,\qquad y\ge0$$

The curve

$$x^{\frac23}+y^{\frac23}=1$$

is an astroid.

Total area of the astroid is

$$\frac{3\pi}{8}$$

By symmetry, area in each quadrant is

$$\frac14\times\frac{3\pi}{8}=\frac{3\pi}{32}$$

Now,

$$y\ge0$$

restricts the region to the upper half-plane.

Also,

$$x+y\ge0$$

represents the region above the line

$$y=-x$$

In the first quadrant, this condition is always satisfied.

Hence, complete first quadrant area is included:

$$\frac{3\pi}{32}$$

In the second quadrant, the line

$$y=-x$$

bisects the astroid region symmetrically.

Therefore, only half of the second quadrant area is included:

$$\frac12\times\frac{3\pi}{32}=\frac{3\pi}{64}$$

Thus,

$$A=\frac{3\pi}{32}+\frac{3\pi}{64}=\frac{9\pi}{64}$$

Therefore,

$$\frac{256A}{\pi}=\frac{256}{\pi}\times\frac{9\pi}{64}=36$$

Hence,

$$\boxed{36}$$