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Understanding divisibility is essential when trying to determine if one number can be evenly distributed into another without any remainder. In general, a number 'a' is divisible by another number 'b' if, upon division, the result is a whole number without any remainder. The relation of divisibility is fundamental in arithmetic and number theory, often used to simplify problems and find patterns within sets of numbers.

For example, in the textbook exercise, we are interested in whether the expression involving consecutive cube numbers \( n^3 + (n+1)^3 + (n+2)^3 \) is divisible by 9. Proving this divisibility can demonstrate a deeper pattern in the behavior of cube numbers, which could extend to various applications in mathematics.

Natural numbers are the set of all positive integers starting from 1, extending to infinity. These are the numbers we typically use to count objects in the real world and they form the basis of many mathematical concepts. In our exercise, we are concerned with proving a property for all natural numbers \( n \), indicating a universal truth in the realm of tallying and quantifying objects.

The set of natural numbers, denoted by \( \mathbb{N} \), includes numbers like 1, 2, 3, and so forth. Such a set is infinite and is considered 'discrete' because there are distinct 'steps' from one number to the next without fractions or decimals in between.

When employing mathematical induction, the inductive hypothesis serves as a pivotal assumption. It's the belief that a certain property holds true for a particular but arbitrary case, usually denoted by 'k'. In the context of our textbook exercise, by assuming that \( k^3 + (k+1)^3 + (k+2)^3 \) is divisible by 9, we take a crucial leap of faith that sets the foundation for proving the property for all cases.

This assumption is not picked out of thin air; it's based on checking an initial case, known as the base case, and then presuming this case holds true for a generic step 'k'. We don't prove it just yet; we utilize it in our logic to demonstrate the property for 'k+1', thereby inching further in our proof.

The inductive step is where the mathematical magic happens in a proof by induction. It is here that we move from the assumption — our inductive hypothesis — to a general proof. In plain terms, if the property is true for 'k', we show it must also be true for 'k+1'. In our specific exercise, we demonstrate that given the expression is divisible by 9 for 'k', it must follow for 'k+1' by including the terms \((k+1)^3+(k+2)^3+(k+3)^3\) and accounting for the transition with the addition of 3k + 9.

This step relies on algebraic manipulation and logical reasoning to bridge the gap between the assumed and the proven. Once this step is established, we've got all the pieces we need to finalize the proof for all natural numbers, thus showcasing the power of mathematical induction as a tool for proving infinite cases with a finite process.