Solve the following equation:

$$
\sqrt[3]{x}+\sqrt[3]{1-x}=2
$$
Solve the following equation:

$$
\sqrt[3]{x}+\sqrt[3]{1-x}=2
$$

Solution: Cube both sides. If we use the identity

$$
(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)
$$

on the left side, then after rearranging, our equation becomes:

$$
3 \cdot \sqrt[3]{x} \cdot \sqrt[3]{1-x} \cdot(\sqrt[3]{x}+\sqrt[3]{1-x})=7
$$

Notice that the expression in parentheses is precisely the left side of (1), which is 2 according to the condition. Thus,

$$
3 \cdot \sqrt[3]{x} \cdot \sqrt[3]{1-x} \cdot 2=7, \quad \text { from which }
$$

$$
\sqrt[3]{x(1-x)}=\frac{7}{6}
$$

If $x>1$, then $x(1-x)<0$, and if $0 \leqq x \leqq 1$, then $1 \geqq 1-x \geqq 0$ also holds, so $x(1-x)<1$. Therefore, on the left side, we always have the cube root of a number less than 1, which is also less than 1. Thus, the equation has no solution in the real numbers.

Remark. If we cube (3), then after rearranging, we arrive at the equation $x^{2}-x+\left(\frac{7}{6}\right)^{3}=0$. The discriminant of this equation is negative, so the equation has no real solution.