Let be an odd natural number $ n\ge 3. $ Find all continuous functions $ f:[0,1]\longrightarrow\mathbb{R} $ that satisfy the following equalities.
$$ \int_0^1 \left( f\left(\sqrt[k]{x}\right) \right)^{n-k} dx=k/n,\quad\forall k\in\{ 1,2,\ldots ,n-1\} $$


[i]Titu Andreescu[/i]
1. **Substitution and Reformulation:**
   Given the integral equation:
   \[
   \int_0^1 \left( f\left(\sqrt[k]{x}\right) \right)^{n-k} dx = \frac{k}{n}, \quad \forall k \in \{1, 2, \ldots, n-1\}
   \]
   We perform the substitution \( x = t^k \). Then \( dx = k t^{k-1} dt \). The limits of integration remain the same since \( t \) ranges from 0 to 1. Thus, the integral becomes:
   \[
   \int_0^1 f(t)^{n-k} t^{k-1} k t^{k-1} dt = \frac{k}{n}
   \]
   Simplifying, we get:
   \[
   k \int_0^1 f(t)^{n-k} t^{2k-2} dt = \frac{k}{n}
   \]
   Dividing both sides by \( k \):
   \[
   \int_0^1 f(t)^{n-k} t^{2k-2} dt = \frac{1}{n}
   \]

2. **Considering the Integral for Different \( k \):**
   This equation must hold for all \( k \in \{1, 2, \ldots, n-1\} \). Specifically, we have:
   \[
   \int_0^1 f(t)^{n-1} dt = \frac{1}{n}
   \]
   \[
   \int_0^1 f(t)^{n-2} t^2 dt = \frac{1}{n}
   \]
   \[
   \int_0^1 f(t)^{n-3} t^4 dt = \frac{1}{n}
   \]
   and so on.

3. **Using Cauchy-Schwarz Inequality:**
   Consider the Cauchy-Schwarz inequality in the context of the integrals:
   \[
   \left( \int_0^1 f(t)^{n-k} t^{2k-2} dt \right)^2 \leq \left( \int_0^1 f(t)^{2(n-k)} dt \right) \left( \int_0^1 t^{4k-4} dt \right)
   \]
   Given that:
   \[
   \int_0^1 t^{4k-4} dt = \frac{1}{4k-3}
   \]
   and:
   \[
   \int_0^1 f(t)^{2(n-k)} dt
   \]
   We need to show that equality holds in the Cauchy-Schwarz inequality, which implies that \( f(t) \) must be proportional to \( t \).

4. **Conclusion by Continuity:**
   Since \( f \) is continuous and the equality holds for all \( k \), we conclude that \( f(t) = t \) satisfies the given integral equation. Substituting \( f(t) = t \) into the original integral:
   \[
   \int_0^1 t^{n-k} t^{k-1} dt = \int_0^1 t^{n-1} dt = \frac{1}{n}
   \]
   This confirms that \( f(t) = t \) is indeed a solution.

\(\blacksquare\)

The final answer is \( \boxed{ f(t) = t } \).