Find all non-negative real numbers $a$ such that the equation \[ \sqrt[3]{1+x}+\sqrt[3]{1-x}=a \] has at least one real solution $x$ with $0 \leq x \leq 1$.
For all such $a$, what is $x$?
1. Consider the given equation:
   \[
   \sqrt[3]{1+x} + \sqrt[3]{1-x} = a
   \]
   We need to find all non-negative real numbers \(a\) such that this equation has at least one real solution \(x\) with \(0 \leq x \leq 1\).

2. Define the function \(f(x)\) as:
   \[
   f(x) = \sqrt[3]{1+x} + \sqrt[3]{1-x}
   \]
   We need to analyze the behavior of \(f(x)\) over the interval \([0, 1]\).

3. Evaluate \(f(x)\) at the endpoints of the interval:
   \[
   f(0) = \sqrt[3]{1+0} + \sqrt[3]{1-0} = \sqrt[3]{1} + \sqrt[3]{1} = 1 + 1 = 2
   \]
   \[
   f(1) = \sqrt[3]{1+1} + \sqrt[3]{1-1} = \sqrt[3]{2} + \sqrt[3]{0} = \sqrt[3]{2} + 0 = \sqrt[3]{2}
   \]

4. Since \(f(x)\) is a sum of two cube root functions, both of which are continuous and differentiable, \(f(x)\) is also continuous and differentiable on \([0, 1]\).

5. To determine the monotonicity of \(f(x)\), compute its derivative:
   \[
   f'(x) = \frac{d}{dx} \left( \sqrt[3]{1+x} + \sqrt[3]{1-x} \right)
   \]
   Using the chain rule:
   \[
   f'(x) = \frac{1}{3}(1+x)^{-\frac{2}{3}} - \frac{1}{3}(1-x)^{-\frac{2}{3}}
   \]

6. Analyze the sign of \(f'(x)\) on \([0, 1]\):
   \[
   f'(x) = \frac{1}{3}(1+x)^{-\frac{2}{3}} - \frac{1}{3}(1-x)^{-\frac{2}{3}}
   \]
   For \(0 \leq x \leq 1\), \((1+x)^{-\frac{2}{3}} \leq (1-x)^{-\frac{2}{3}}\), hence \(f'(x) \leq 0\).

7. Since \(f'(x) \leq 0\) on \([0, 1]\), \(f(x)\) is a decreasing function on this interval.

8. Given that \(f(x)\) is decreasing on \([0, 1]\), the range of \(f(x)\) on this interval is:
   \[
   [\sqrt[3]{2}, 2]
   \]

9. Therefore, the values of \(a\) for which the equation has at least one solution \(x\) in \([0, 1]\) are:
   \[
   a \in [\sqrt[3]{2}, 2]
   \]

10. For each \(a \in [\sqrt[3]{2}, 2]\), the corresponding \(x\) can be found by solving:
    \[
    x = \sqrt{1 - \left( \frac{a^3 - 2}{3a} \right)^3}
    \]

The final answer is \( \boxed{ a \in [\sqrt[3]{2}, 2] } \) and \(x = \sqrt{1 - \left( \frac{a^3 - 2}{3a} \right)^3}\)