Duality

Observed that there is symmetricality (because it is inscribed in a cube). So the octahedron can be divided into two tetrahedron (a tetrahedron is a pyramid). So what we need to know is the edge length of the tetrahedron. To find this, we connect the midpoints of the square (see my figure). We form another square. This is the base of the tetrahedron (middle part of the octahedron). By Pythagorean theorem, the length is $\sqrt{2}$ . Now the height of each tetrahedron is $\dfrac{2}{2}=1$ . So the desired volume is,

$V=2\left(\dfrac{1}{3}\right)(\sqrt{2})^2(1)=\dfrac{4}{3}$ .

Note:

The side length of the cube can be computed using the formula, $V=a^3$ , where $V$ is the volume and $a$ is the edge length. The edge length is $2$ .

The cube's side is $\sqrt{8 \text{in}^3}=2 \text{in}$

Let cut the octahedron in two squared pyramids of shared base. Its base is a square formed by join the centers of four adjacent faces of the cube so that the diagonal of the base is $2 \text{in} \therefore$ its area is $\frac{2\times2}{2}=2$ (remember: a square is also a rhombus with equal diagonal), also it height is a half of the square's side that is $1 \text{in}$ . Pyramid's volume is $\frac{b\times h}{3}\therefore$ the octahedron's volume is $2\times\frac{2\times1}{3}=\boxed{\frac{4}{3}\text{in}^3}$

Cube:

v = s³

8 = s³

s = 2 inches

Octahedron:

v = $\frac{1}{3}$ $x^3$ $\sqrt{2}$

from the figure

x = $\sqrt{2}$

v = $\frac{1}{3}$ $*($ $\sqrt{2})^3$ $*$ $\sqrt{2}$

v = $\frac{4}{3}$ $in^3$

It is not complicated. First, we need to know that a square within a square will have an area of a half of the original square. A pyramid will have a volume one third of the prisim.

Applying these knowledge, we know: A = 8 x 0.5 x (1/3) = (4/3)

That's it.

V=2((1/3)sqrt{2})^2 = (2/3)(2) = 4/3 in^{3}