Circle buddies enclosed in a square

Let the width of the green triangle be $a$ and its height $2b$ as shown in the figure. Note that the central red circle and the two semicircles have the same radius of $\frac 12$ . The two semicircles touch at the center of the red circle. The center of the red circle and its two intersecting points with the semicircles form an equilateral triangle with side length $a$ equal to the radius of the red circle which is $\frac 12$ . That is $a=\frac 12$ . The center of the red circle, one of its intersecting point with a semicirle and the point it touches the square also form an equilateral triangle with side length $\frac 12$ . Then $b = a \sin 60^\circ = \frac {\sqrt 3}4$ . Therefore the area of the green rectangle is $A = 2ab = 2 \times \frac 12 \times \frac {\sqrt 3}4 = \frac {\sqrt 3}4$ .

Therefore, $(b^3-a^3)^2 = (4^3-3^2)^2 = (64-27)^2 = 37^2 = \boxed{1369}$ .

Note: All angles in this solution are in degrees, not radians.

We can make 2 equilateral triangles on each side of this rectangle with side length $\frac{1}{2}$ as shown:

Phew! Triangles!

We can use the cosine rule here in order to find the left and right sides of this rectangle: $(\text{Left/Right side})^2 = 0.5^2 + 0.5^2 - 2(0.5)(0.5)\cos(120) = \frac{3}{4}$ $\text{Left/Right side} = \sqrt{\frac{3}{4}} = \boxed{\frac{\sqrt{3}}{2}}$

We can see that the diagonal length inside the rectangle is 1 because it is the inner circle's diameter. Now we can use the Pythagorean Theorem to find the lower and upper sides: $\text{Lower/upper sides} = \sqrt{1^2-\left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4}} = \boxed{\frac{1}{2}}$

Now we can calculate the area: $\text{Area}=\frac{1}{2}\cdot \frac{\sqrt{3}}{2}=\boxed{\frac{\sqrt{3}}{4}}$

We can see here that $a=3$ and $b=4$ . The side length of the square that we need to find the area of is $4^3-3^3 = 37$ and the area is $\large 37^2 = \green{\boxed{1369}}$