Let $a \geq 0$ be a natural number. Determine all rational $x$, so that
\[\sqrt{1+(a-1)\sqrt[3]x}=\sqrt{1+(a-1)\sqrt x}\]
All occurring square roots, are not negative.

[b]Note.[/b] It seems the set of natural numbers = $\mathbb N = \{0,1,2,\ldots\}$ in this problem.
1. **Case when \(a = 1\):**
   \[
   \sqrt{1 + (1-1)\sqrt[3]{x}} = \sqrt{1 + (1-1)\sqrt{x}}
   \]
   Simplifying both sides:
   \[
   \sqrt{1 + 0} = \sqrt{1 + 0} \implies \sqrt{1} = \sqrt{1} \implies 1 = 1
   \]
   This equality holds for all \(x \in \mathbb{R}\). Therefore, for \(a = 1\), any \(x\) is a solution.

2. **Case when \(a \neq 1\):**
   Given the equation:
   \[
   \sqrt{1 + (a-1)\sqrt[3]{x}} = \sqrt{1 + (a-1)\sqrt{x}}
   \]
   Squaring both sides to eliminate the square roots:
   \[
   1 + (a-1)\sqrt[3]{x} = 1 + (a-1)\sqrt{x}
   \]
   Subtracting 1 from both sides:
   \[
   (a-1)\sqrt[3]{x} = (a-1)\sqrt{x}
   \]
   Since \(a \neq 1\), we can divide both sides by \(a-1\):
   \[
   \sqrt[3]{x} = \sqrt{x}
   \]
   Cubing both sides to eliminate the cube root:
   \[
   (\sqrt[3]{x})^3 = (\sqrt{x})^3 \implies x = x^{3/2}
   \]
   Simplifying the equation:
   \[
   x = x^{3/2}
   \]
   Raising both sides to the power of 2/3:
   \[
   x^{2/3} = x
   \]
   Rearranging the equation:
   \[
   x^{2/3} - x = 0 \implies x(x^{2/3 - 1}) = 0 \implies x(x^{-1/3}) = 0 \implies x^{2/3} = 0
   \]
   This implies:
   \[
   x = 0 \quad \text{or} \quad x^{2/3} = 1 \implies x = 1
   \]
   Therefore, the solutions are \(x = 0\) and \(x = 1\).

3. **Verification:**
   - For \(x = 0\):
     \[
     \sqrt{1 + (a-1)\sqrt[3]{0}} = \sqrt{1 + (a-1)\sqrt{0}} \implies \sqrt{1 + 0} = \sqrt{1 + 0} \implies \sqrt{1} = \sqrt{1} \implies 1 = 1
     \]
   - For \(x = 1\):
     \[
     \sqrt{1 + (a-1)\sqrt[3]{1}} = \sqrt{1 + (a-1)\sqrt{1}} \implies \sqrt{1 + (a-1)} = \sqrt{1 + (a-1)} \implies \sqrt{a} = \sqrt{a}
     \]
   Both values satisfy the original equation.

The final answer is \(\boxed{x = 0, 1}\)