It is often easier to first draw a diagram for such a problem.

2011AMC10B22.png
Sometimes, it may also be easier to think of the problem in 2D. Take a cross section of the pyramid through the apex and two points from the base that are opposite to each other. Place it in two dimensions.

unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(10pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red);

The dimensions of this triangle are $1, 1,$ and $\sqrt{2}$ because the sidelengths of the pyramid are $1$ and the base of the triangle is the diagonal of the pyramid's base. This is a $45-45-90$ triangle. Also, we can let the dimensions of the rectangle be $s$ and $s\sqrt{2}$ because the longer length was the diagonal of the cube's base and the shorter length was a side of the cube.

The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$

Now we can just use segment addition to find the value of $s.$
$$\sqrt{2}=s+s\sqrt{2}+s=2s+s\sqrt{2}=(2+\sqrt{2})s$$
$$s=\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$$
The volume of the cube is $s^3 = (\sqrt{2}-1)^3 = (\sqrt{2}-1)(3-2\sqrt{2}) = 3\sqrt{2}-3-4+2\sqrt{2} = \boxed{\textbf{(A)} 5\sqrt{2}-7}$