403*. Solve the equation:

$$
\sqrt[4]{1-x^{2}}+\sqrt[4]{1-x}+\sqrt[4]{1+x}=3
$$
$\triangle$ The domain of the equation is the interval $[-1 ; 1]$. In this domain, we can apply the inequality between the geometric mean and the arithmetic mean of two non-negative numbers to each of the radicals in the left-hand side:

$\sqrt[4]{1-x^{2}}=\sqrt{\sqrt{1+x} \cdot \sqrt{1-x}} \leq \frac{\sqrt{1+x}+\sqrt{1-x}}{2}$,

$\sqrt[4]{1+x}=\sqrt{1 \cdot \sqrt{1+x}} \leq \frac{1+\sqrt{1+x}}{2}$,

$\sqrt[4]{1-x}=\sqrt{1 \cdot \sqrt{1-x}} \leq \frac{1+\sqrt{1-x}}{2}$.

We will add these three inequalities term by term, and apply the same inequality twice more to the expression in the right-hand side of the new inequality:

$$
\sqrt[4]{1-x^{2}}+\sqrt[4]{1+x}+4 \sqrt[4]{1-x} \leq 1+\sqrt{1+x}+\sqrt{1-x} \leq 1+\frac{1+(1+x)}{2}+\frac{1+(1-x)}{2}=3
$$

The obtained inequality must turn into an equality according to the condition. However, this is only possible if each of the inequalities used here turns into an equality. It is known that the geometric mean of two non-negative numbers equals their arithmetic mean only when these numbers are equal. In this case, we will have:

$$
1=1+x=1-x
$$

from which $x=0$.

Answer: 0.