10.5. Find all pairs $(x ; y)$ of real numbers that satisfy the conditions: $x^{3}+y^{3}=1$ and $x^{4}+y^{4}=1$.
Answer: $(0 ; 1) ;(1 ; 0)$.

Solution. First method. From the second equality, it follows that $|x| \leq 1$ and $|y| \leq 1$. For the first equality to hold, at least one of the numbers $x$ or $y$ must be positive. Since both equalities are symmetric with respect to the variables, let $0 < x \leq 1$. Consider the function $f(t)=t^{4}+(1-t)^{4}-1$. We have $f(0)=f(1)=0$. If $0 < t < 1$, then $f(t) > 0$. If $t > 1$, then $t^{4} > 1$ and $t^{4} + (1-t)^{4} > 1$. If $t < 0$, then $t^{4} > 1$ and $t^{4} + (1-t)^{4} > 1$.

Therefore, $f(t)=0 \Leftrightarrow t=0$ or $t=1$. Substituting these values into the original equalities instead of any of the variables, we get the answer.

Evaluation criteria.

“+” A complete and well-reasoned solution is provided

“士” A generally correct reasoning is provided, containing minor gaps or inaccuracies (e.g., the symmetric answer is not considered)

“Ғ” There are correct ideas for estimating the values of the variables, but the solution is not fully developed

“-” Only the answer is provided

“-” The problem is not solved or is solved incorrectly