HELP PLZ!!!!

Let  a  be the length of an edge of the tetrahedron.

A face of the cube looks like this:

By the Pythagorean theorem,

6² + 6²  =  a²

                            Combine like terms....  n + n = n * 2    so    6² + 6²  =  6² * 2

6² * 2  =  a²

                            Take the positive square root of both sides of the equation.

√[ 6² * 2 ]  =  a

√6²  *  √2  =  a

6√2  =  a

Since each face of the cube is the same, each edge of the tetrahedron is the same.

So the tetrahedron is a regular tetrahedron.

The formula for the volume of a regular tetrahedron is

volume  =  (edge length)³ / ( 6√2 )

                                                                   Plug in  6√2  for the edge length.

volume  =  ( 6√2 )³ / ( 6√2 )

                                                                   Simplify.

volume  =  ( 6√2 )( 6√2 )( 6√2 ) / ( 6√2 )

volume  =  ( 6√2 )( 6√2 )

volume  =  36 * 2

volume  =  72      cubic units

A cube has side 6 lengths. Its vertices are alternately colored black and purple, as shown below.
What is the volume of the tetrahedron whose corners are the purple vertices of the cube?
(A tetrahedron is a pyramid with a triangular base.)

A cube has side 6 lengths. $\mathbf{\text{Let $s = 6$}}$

The cube has volume $\mathbf{V_{\text{cube}} = s^3}$

The 4 right triangular pyramids that must be carved off the cube to produce the regular tetrahedron each have volume

$\mathbf{V_{\text{right triangular pyramid}} = \frac13\ \times \frac{s^2}{2} \times s }$

The volume of the regular tetrahedron is  $\mathbf{ V_{\text{cube} }- 4\times V_{\text{right triangular pyramid}} }$

$\begin{array}{|rcll|} \hline && V_{\text{tetrahedron}} \\\\ &=& V_{\text{cube} }- 4\times V_{\text{right triangular pyramid}} \\\\ &=& s^3 - 4\times \dfrac13\ \times \dfrac{s^2}{2} \times s \\\\ &=& s^3 - \dfrac23 s^3 \\\\ &=& \dfrac13 s^3 \quad & | \quad s = 6 \\\\ &=& \dfrac{6^3}{3} \\\\ &=& \dfrac{216}{3} \\\\ &=& 72 \\ \hline \end{array}$

The volume of the tetrahedron is 72