The image contains two mathematical problems in Vietnamese. Here are their translations and explanations:

Problem 1.17:

Given $f(x)$ is a polynomial of one variable.
Prove that if $(x - 1) \mid f(x^n)$,
then $(x^n - 1) \mid f(x^n)$.

Explanation:
To prove this, you need to use the relationship between divisors and roots of polynomials. Start by noting that if $(x - 1) \mid f(x^n)$, it means $x = 1$ is a root of $f(x^n)$. From this, deduce how the factor $x^n - 1$ relates to $f(x^n)$ because $x^n - 1$ represents all $n$-th roots of unity.

Problem 1.18:

Given two polynomials $f(x)$ and $g(x)$.
Prove that if $x^2 + x + 1 \mid \big( f(x^3) + xg(x^3) \big)$,
then $x - 1 \mid f(x)$ and $x - 1 \mid g(x)$.

Explanation:
This involves factoring and using properties of divisors of polynomials. The condition $x^2 + x + 1 \mid f(x^3) + xg(x^3)$ implies $x^2 + x + 1$, which contains the cube roots of unity $\omega$, divides the given expression. From this, investigate the behavior of $f(x)$ and $g(x)$ under substitution and the factorization of $x^3 - 1$ into $(x - 1)(x^2 + x + 1)$.

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Related Questions:

1. What is the significance of roots of unity in these problems?
2. How can divisors of a polynomial help prove such divisibility conditions?
3. What role does $x^3 - 1$ play in Problem 1.18?
4. Can these ideas be extended to general cases of $f(x^n)$?
5. What are practical applications of these polynomial divisibility problems?

Tip:

Always factorize complex polynomials into irreducible components like $x^n - 1 = (x - 1)(x^{n-1} + \ldots + x + 1)$ to simplify divisibility proofs.