\section*{Problem 3 - 071243}

Give all functions \(y=f(x)\) that satisfy the equation

\[
a \cdot f\left(x^{n}\right)+f\left(-x^{n}\right)=b x
\]

in the largest possible domain (within the range of real numbers), where \(b\) is any real number, \(n\) is any odd natural number, and \(a\) is a real number with \(|a| \neq 1\)!
}

Since \(n\) is odd, the identity \((-x)^{n}=-x^{n}\) holds for all real \(x\). In particular, by substituting \(-x\) into the functional equation, one obtains a second equation:

\[
a \cdot f\left(-x^{n}\right)+f\left(x^{n}\right)=-b x
\]

Adding these two equations yields \((a+1) \cdot\left(f\left(x^{n}\right)+f\left(-x^{n}\right)\right)=0\), and thus, because \(a \neq-1\), we have \(f\left(-x^{n}\right)=-f\left(x^{n}\right)\). Substituting this back into the original functional equation, we get \((a-1) \cdot f\left(x^{n}\right)=b x\), and after dividing by \(a-1 \neq 0\) and the appropriate substitution,

\[
f(x)=\frac{b}{a-1} \cdot \sqrt[n]{x}
\]

One can quickly verify that this function indeed satisfies the functional equation, since

\[
\begin{gathered}
f\left(x^{n}\right)=\frac{b}{a-1} \cdot \sqrt[n]{x^{n}}=\frac{b}{a-1} \cdot x \quad \text { and } \\
f\left(-x^{n}\right)=\frac{b}{a-1} \cdot \sqrt[n]{-x^{n}}=\frac{b}{a-1} \cdot \sqrt[n]{(-x)^{n}}=\frac{b}{a-1} \cdot(-x)
\end{gathered}
\]

thus,

\[
a \cdot f\left(x^{n}\right)+f\left(-x^{n}\right)=a \cdot \frac{b}{a-1} \cdot x-\frac{b}{a-1} \cdot x=(a-1) \cdot \frac{b}{a-1} \cdot x=b x
\]

Remark: For this function to be well-defined, for negative real numbers \(x\) and odd root exponents \(n\), the root function must be extended to the entire set of real numbers via \(\sqrt[n]{x}=-\sqrt[n]{-x}\), so that it serves as the inverse function of the power function \(p: x \mapsto x^{n}\) over the entire set of real numbers.

This is possible because \(p\) is a one-to-one mapping from the real numbers to the real numbers.