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Its volume would be multiplied by the cube of 2 and become 8 m 3. The original cube (1 m sides) has a surface area to volume ratio of 6:1. The larger (2 m sides) cube has a surface area to volume ratio of (24/8) 3:1. As the dimensions increase, the volume will continue to grow faster than the surface area. Thus the square–cube law.
Proof without words of the Nicomachus theorem (Gulley (2010)) that the sum of the first n cubes is the square of the n th triangular number. In mathematics, a proof without words (or visual proof) is an illustration of an identity or mathematical statement which can be demonstrated as self-evident by a diagram without any accompanying explanatory text.
"Completing the square" consists to remark that the two first terms of a quadratic polynomial are also the first terms of the square of a linear polynomial, and to use this for expressing the quadratic polynomial as the sum of a square and a constant. Completing the cube is a similar technique that allows to transform a cubic polynomial into a ...
A square whose side length is a triangular number can be partitioned into squares and half-squares whose areas add to cubes. From Gulley (2010).The n th coloured region shows n squares of dimension n by n (the rectangle is 1 evenly divided square), hence the area of the n th region is n times n × n.
Another geometric proof proceeds as follows: We start with the figure shown in the first diagram below, a large square with a smaller square removed from it. The side of the entire square is a, and the side of the small removed square is b. The area of the shaded region is . A cut is made, splitting the region into two rectangular pieces, as ...
The cube of a number or any other mathematical expression is denoted by a superscript 3, for example 2 3 = 8 or (x + 1) 3. The cube is also the number multiplied by its square: n 3 = n × n 2 = n × n × n. The cube function is the function x ↦ x 3 (often denoted y = x 3) that maps a number to its cube. It is an odd function, as
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Proof by exhaustion can be used to prove that if an integer is a perfect cube, then it must be either a multiple of 9, 1 more than a multiple of 9, or 1 less than a multiple of 9. [3] Proof: Each perfect cube is the cube of some integer n, where n is either a multiple of 3, 1 more than a multiple of 3, or 1 less than a multiple of 3. So these ...
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