Question 1200656
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Answer: <font color=red size=4> V = (d+1)^3 </font>


That expands out to
V = d^3 + 3d^2 + 3d + 1


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Work Shown:


d = diameter of the ball
d+1 = one inch larger than that diameter = side length of the cube


V = volume of the cube
V = (side)*(side)*(side)
V = (side)^3
<font color=red size=4> V = (d+1)^3 </font>


Optionally we can expand that out
V = (d+1)^3
V = (d+1)(d+1)^2
V = (d+1)(d^2+2d+1)
V = d(d^2+2d+1)+1(d^2+2d+1)
V = d^3+2d^2+d+d^2+2d+1
V = d^3+(2d^2+d^2)+(d+2d)+1
V = d^3+3d^2+3d+1


The <a href = "https://en.wikipedia.org/wiki/Binomial_theorem">Binomial Theorem</a> is another approach you could take to go from (d+1)^3 to d^3+3d^2+3d+1
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