For this question, the student will have to use Trial and error
via a table diagram.
Let the amount of marbles on one side be A, and let the total
amount of marbles be T.
In short, T = A X A X A
l__A__l__6__l__7__l__8__l
l__T__l_216_l_343_l_512_l
When A = 6, T = 6 X 6 X 6 = 216
When A = 7, T = 7 X 7 X 7 = 343
When A = 8, T = 8 X 8 X 8 = 512
Thus, there are 8 marbles on
each side of the square.
What???
A square has 4 sides. Why are you cube rooting 512 to get 8?
Even IF your method was correct (thought I doubt it), it should
be more of x^4 = 512.
8 + 8 + 8 + 8 is wrong.
8 x 8 x 8 x 8 is wrong.
8 + 8 + 8 is wrong.
8 x 8 x 8 is wrong too.
Answer should be 512 minus 4, followed by divided by 4, and then
plus 2.
Answer is 129.
Explaination below.
Y ............X..............Y
.
.
.
X X
.
.
.
Y ............X..............Y
Look carefully.
Y = The marble at each corner of the square = 1
Minus out the 4Y, which are the marbles at all four
corners.
512 - 4Y = 512 - 4 = 508
This gives you 4X.
4X = 508
Then divide 4X by 4. This gives you X.
X = 127
Then add X + Y + Y to get the total marbles on one side. The
answer will be the same for all sides of the square.
127 + 1 + 1 = 129
Correct me if I'm wrong, but I am 100% sure the solution
provided by ForbiddenSinner is wrong.
Therefore answer = 129.