Suppose that a digitized TV picture is to be transmitted from a source that uses a matrix of 480 * 500 picture elements (pixels)
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The volume of a box with dimensions x by x by x is 
.
So if we are given the volume V of a cube-shaped box, the side length of thit is![\sqrt[3]{V}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7BV%7D%20)
.
So we calculate the cubic roots of the volumes we have, and we add their heights.
To calculate the cubic root of 729, we can factorize it, and group the perfect cubes together, as follows:
, which we recognize as 
so
similarly 1,331 can be found to be
.
Thus we have 2 boxes with side length equal to 11 m and one with side length equal to 9 m.
Answer: h= 11+11+9 = 31 (meters)
So if we are given the volume V of a cube-shaped box, the side length of thit is
So we calculate the cubic roots of the volumes we have, and we add their heights.
To calculate the cubic root of 729, we can factorize it, and group the perfect cubes together, as follows:
so
similarly 1,331 can be found to be
Thus we have 2 boxes with side length equal to 11 m and one with side length equal to 9 m.
Answer: h= 11+11+9 = 31 (meters)