251. Find all real solutions of the system

$$
\left\{\begin{array}{c}
x^{3}+y^{3}=1 \\
x^{4}+y^{4}=1
\end{array}\right.
$$
251. If $x^{4}+y^{4}=1$, then either $x^{4}=1, y^{4}=0$, or $x^{4}=0$, $y^{4}=1$, or, finally, $0x^{4}+y^{4}=1
\end{gathered}
$$

from which it follows that two numbers $x, y$ of the same sign, such that $|x|<1$, $|y|<1$, cannot serve as a solution to our system. It is even clearer that
two numbers $x, y$ of different signs, where $|x|<1,|y|<1$, also cannot serve as a solution, in this case

$$
x^{3}+y^{3} \leqslant\left|x^{3}+y^{3}\right|<\max \left(\left|x^{3}\right|,\left|y^{3}\right|\right)<1
$$

Thus, the solutions to the system must be sought among such values of $x, y$ that $x^{4}=1, y^{4}=0$, or vice versa, i.e., among the pairs $x= \pm 1, y=0$ and $x=0, y= \pm 1$; it is clear that of these four pairs of numbers, only two are solutions: $x=1, y=0$ and $x=0, y=1$.