A cube, an octahedron, and a volume

Let the vertices of the cube have coordinates of $(\pm 3, \pm 3, \pm 3)$ , so that the vertices of the yellow triangle are $(3, 0, 0)$ , $(0, 3, 0)$ , and $(0, 0, 3)$ .

Then the equation of the plane containing the yellow triangle is $x + y + z = 3$ , which passes through three of the cube's vertices $(3, 3, -3)$ , $(3, -3, 3)$ , and $(-3, 3, 3)$ .

Therefore, the volume of the portion of the cube that lies above this plane is a right-angled tetrahedron, which has a volume of $\frac{1}{6}6^3 = \boxed{36}$ .

line the 3 adjacent vertexes of top-right vertex of front face, then get an equilateral triangle, which is the same plane of yellow triangle, as the vertexes of yellow triangle are at the midpoint of sides of it respectively.