1. **Problem statement:** We want to determine if for all integers $n$ and $m$, the condition "if $n - m$ is even, then $n^3 - m^3$ is even" is true. 2. **Recall definitions and properties:** - An integer is even if it can be written as $2k$ for some integer $k$. - The difference $n - m$ is even means $n - m = 2k$ for some integer $k$. 3. **Express $n$ in terms of $m$:** Since $n - m = 2k$, we have $n = m + 2k$. 4. **Calculate $n^3 - m^3$ using the factorization formula:** $$n^3 - m^3 = (n - m)(n^2 + nm + m^2)$$ 5. **Substitute $n - m = 2k$:** $$n^3 - m^3 = 2k (n^2 + nm + m^2)$$ 6. **Analyze the parity:** Since $2k$ is even, the product $2k (n^2 + nm + m^2)$ is even regardless of the value of $(n^2 + nm + m^2)$ because an even number times any integer is even. 7. **Conclusion:** Therefore, if $n - m$ is even, then $n^3 - m^3$ is also even. The statement is true.