The answer is 114sqrt{6} in³
A regular hexahedron is actually a cube.
Diagonal of a cube D is a hypotenuse of a right triangle which other two legs are face diagonal (f) and length of a side (a):
D² = f² + a²
Face diagonal is a hypotenuse of a right triangle which sides are a and a:
f² = a² + a² =2a²
D² = f² + a²
f² = 2a²
D² = 2a² + a² = 3a²
D = √3a² = √3 * √a² = √3 * a = a√3
Volume of a cube with side a is: V = a³
D = a√3
⇒ a = D/√3
V = a³ = (D/√3)³
We have:
D = 8√2 in
[tex]V= (\frac{D}{ \sqrt{3} } )^{3} =( \frac{8 \sqrt{2} }{ \sqrt{3} } )^{3}=( \frac{8 \sqrt{2} * \sqrt{3} }{ \sqrt{3}* \sqrt{3}} )^{3} =( \frac{8 \sqrt{2*3} }{ \sqrt{3*3}} )^{3} =( \frac{8 \sqrt{6} }{ \sqrt{3^{2} }} )^{3} =( \frac{8 \sqrt{6} }{ 3} )^{3} = \\ \\ = \frac{ 8^{3} *( \sqrt{6} )^{3}}{3^{3}} = \frac{512* \sqrt{6 ^{2} }* \sqrt{6} }{27} = \frac{512*6* \sqrt{6} }{27} = 114sqrt{6} [/tex]
