1. **Problem statement:** Prove algebraically that for all positive integers $n$, the expression $n^3 + 3n^2 - 2n$ is always even. 2. **Formula and rules:** An integer is even if it is divisible by 2. We want to show that $n^3 + 3n^2 - 2n$ is divisible by 2 for all positive integers $n$. 3. **Step 1: Factor the expression** $$n^3 + 3n^2 - 2n = n(n^2 + 3n - 2)$$ 4. **Step 2: Factor the quadratic inside the parentheses** $$n^2 + 3n - 2 = (n + 2)(n - 1)$$ 5. **Step 3: Rewrite the expression** $$n(n + 2)(n - 1)$$ 6. **Step 4: Analyze the product of three consecutive integers** Note that $n - 1$, $n$, and $n + 2$ are not three consecutive integers, but consider the integers $n - 1$, $n$, and $n + 1$ which are consecutive. Since $n + 2 = (n + 1) + 1$, the product $n(n + 2)(n - 1)$ includes two integers separated by 1 and one integer before them. 7. **Step 5: Use parity argument** Among any two consecutive integers, one is even. Here, $n$ and $n + 2$ differ by 2, so they have the same parity. The integer $n - 1$ differs by 1 from $n$, so $n - 1$ and $n$ have opposite parity. 8. **Step 6: Check cases for $n$ even or odd** - If $n$ is even, then $n$ is divisible by 2, so the product is even. - If $n$ is odd, then $n - 1$ is even, so the product is even. 9. **Step 7: Conclusion** Since in both cases the product $n(n + 2)(n - 1)$ is divisible by 2, the expression $n^3 + 3n^2 - 2n$ is always even for all positive integers $n$. **Final answer:** $n^3 + 3n^2 - 2n$ is always even for all positive integers $n$.