To prove that $(n+1)^3 - n^3$ is always odd for all $n \in \mathbb{N}$ using mathematical induction, we follow these steps:

Base Case:

For $n = 1$, we calculate: $(n+1)^3 - n^3 = (1+1)^3 - 1^3 = 2^3 - 1^3 = 8 - 1 = 7$ Since 7 is odd, the statement holds true for the base case $n = 1$.

Inductive Hypothesis:

Assume that for some $k \in \mathbb{N}$, the statement holds true, i.e., $(k+1)^3 - k^3 \text{ is odd}.$ This is our inductive hypothesis.

Inductive Step:

We need to show that if the statement is true for $k$, it must also be true for $k+1$. That is, we need to prove that: $((k+1)+1)^3 - (k+1)^3 \text{ is odd}.$

First, expand both sides: $((k+1)+1)^3 - (k+1)^3 = (k+2)^3 - (k+1)^3.$ Now expand both cubes: $(k+2)^3 = (k+2)(k+2)(k+2) = k^3 + 6k^2 + 12k + 8,$ $(k+1)^3 = (k+1)(k+1)(k+1) = k^3 + 3k^2 + 3k + 1.$ Subtract $(k+1)^3$ from $(k+2)^3$: $(k+2)^3 - (k+1)^3 = (k^3 + 6k^2 + 12k + 8) - (k^3 + 3k^2 + 3k + 1).$ Simplifying this: $= k^3 + 6k^2 + 12k + 8 - k^3 - 3k^2 - 3k - 1$ $= 3k^2 + 9k + 7.$ Thus, $(k+2)^3 - (k+1)^3 = 3k^2 + 9k + 7.$ Notice that $3k^2 + 9k + 7$ is the sum of two terms that are multiples of 3 and a constant 7. Since $3k^2 + 9k$ is always even (it’s divisible by 6), adding 7 makes the whole expression odd.

Conclusion:

By the principle of mathematical induction, $(n+1)^3 - n^3$ is always odd for all $n \in \mathbb{N}$.

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Would you like further details on any step or concept? Here are five related questions to consider:

1. What is the principle of mathematical induction?
2. Can you generalize the proof for any polynomial function?
3. How do you identify if an algebraic expression is odd or even?
4. What are some practical applications of induction in proving odd/even properties?
5. Can you use a different method, like direct factorization, to prove this statement?

Tip: Always check the base case carefully; it's crucial to establishing the foundation for induction!