Question 1180627
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Consider the following diagram, depicting one of the cut-off corners with equilateral triangle ABC and "peak" of the pyramid at O.<br>
{{{drawing(400,400,-5,5,-5,5
,line(0,0,5,0),line(0,0,0,5),line(-3,-3,0,0)
,line(-3,-3,5,0),line(-3,-3,0,5),line(5,0,0,5)
,locate(.3,.5,O),locate(-3,-3,A),locate(5,-.2,B),locate(.5,5,C)
,locate(3,3,6),locate(-2.2,1,6),locate(2,-1.5,6)
)}}}<br>
Each of the triangles AOB, BOC, and COA is an isosceles right triangle with hypotenuse 6cm.  That makes OA=OB=OC=3*sqrt(2).<br>
Then the volume of each cut-off corner can be calculated as the volume of a pyramid with base AOB and height OC.<br>
{{{V=(1/3)(base)(height)}}}<br>
Then of course multiply that by 8 to find the total volume of the cut-off corners.<br>
You should end up with a total volume of 72*sqrt(2).<br>