To prove or disprove the statement $\forall \epsilon \in \mathbb{N}, (\epsilon + 1)^3 - 5\epsilon^2 + 6 \text{ is odd}$, we can analyze the expression step by step using formal mathematical logic and deduction.

Step 1: Expression Breakdown

The given expression is:

$f(\epsilon) = (\epsilon + 1)^3 - 5\epsilon^2 + 6$

First, expand $(\epsilon + 1)^3$:

$(\epsilon + 1)^3 = \epsilon^3 + 3\epsilon^2 + 3\epsilon + 1$

Substitute this back into the original expression:

$f(\epsilon) = (\epsilon^3 + 3\epsilon^2 + 3\epsilon + 1) - 5\epsilon^2 + 6$

Simplify:

$f(\epsilon) = \epsilon^3 + (3\epsilon^2 - 5\epsilon^2) + 3\epsilon + 1 + 6$ $f(\epsilon) = \epsilon^3 - 2\epsilon^2 + 3\epsilon + 7$

Thus, the simplified expression is:

$f(\epsilon) = \epsilon^3 - 2\epsilon^2 + 3\epsilon + 7$

Step 2: Analyzing the Parity of the Expression

To determine if $f(\epsilon)$ is odd for all $\epsilon \in \mathbb{N}$, we analyze its behavior based on whether $\epsilon$ is even or odd.

Case 1: $\epsilon$ is even

Let $\epsilon = 2k$, where $k \in \mathbb{N}$. Substituting $\epsilon = 2k$ into $f(\epsilon)$:

$f(2k) = (2k)^3 - 2(2k)^2 + 3(2k) + 7$ $f(2k) = 8k^3 - 8k^2 + 6k + 7$

Now check the parity (even or odd) of each term:

• $8k^3$ is even,
• $8k^2$ is even,
• $6k$ is even,
• $7$ is odd.

Thus, the sum of an even number of terms plus 7 (an odd number) is odd. Therefore, $f(2k)$ is odd for all even $\epsilon$.

Case 2: $\epsilon$ is odd

Let $\epsilon = 2k + 1$, where $k \in \mathbb{N}$. Substituting $\epsilon = 2k + 1$ into $f(\epsilon)$:

$f(2k + 1) = (2k + 1)^3 - 2(2k + 1)^2 + 3(2k + 1) + 7$

First, expand $(2k + 1)^3$ and $(2k + 1)^2$:

$(2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1$ $(2k + 1)^2 = 4k^2 + 4k + 1$

Now substitute into the expression:

$f(2k + 1) = (8k^3 + 12k^2 + 6k + 1) - 2(4k^2 + 4k + 1) + 3(2k + 1) + 7$ $f(2k + 1) = 8k^3 + 12k^2 + 6k + 1 - (8k^2 + 8k + 2) + (6k + 3) + 7$ $f(2k + 1) = 8k^3 + (12k^2 - 8k^2) + (6k - 8k + 6k) + (1 - 2 + 3 + 7)$ $f(2k + 1) = 8k^3 + 4k^2 + 4k + 9$

Now check the parity of each term:

• $8k^3$ is even,
• $4k^2$ is even,
• $4k$ is even,
• $9$ is odd.

Thus, the sum of an even number of terms plus 9 (an odd number) is odd. Therefore, $f(2k + 1)$ is odd for all odd $\epsilon$.

Step 3: Conclusion

Since $f(\epsilon)$ is odd for both even and odd $\epsilon$, we conclude that:

$\forall \epsilon \in \mathbb{N}, (\epsilon + 1)^3 - 5\epsilon^2 + 6 \text{ is odd}.$

Thus, the statement is true.

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Would you like more details or clarifications?

Here are five related questions:

1. How can we prove if a polynomial is always odd or even?
2. How does the parity of individual terms in a polynomial affect the overall result?
3. Can this method be generalized to other cubic expressions?
4. How can we use modular arithmetic to verify if an expression is odd or even?
5. What other properties can be inferred from analyzing the parity of a polynomial?

Tip: Always check specific cases (like even and odd integers) to better understand the behavior of polynomials in parity-related problems.