23. A function $f$ satisfies $y^{3} f(x)=x^{3} f(y)$ and $f(3) \neq 0$. What is the value of $\frac{f(20)-f(2)}{f(3)}$ ?
A 6
B 20
C 216
D 296
E 2023
Solution
D
By putting $x=20$ and $y=3$ in the equation $y^{3} f(x)=x^{3} f(y)$, we have $27 f(20)=8000 f(3)$. Hence
$$
f(20)=\frac{8000}{27} f(3) .
$$

By putting $x=2$ and $y=3$ in the same equation, we have $27 f(2)=8 f(3)$. Hence
$$
f(2)=\frac{8}{27} f(3) \text {. }
$$

Therefore,
$$
\begin{aligned}
f(20)-f(2) & =\frac{8000}{27} f(3)-\frac{8}{27} f(3) \\
& =\left(\frac{8000}{27}-\frac{8}{27}\right) f(3) \\
& =\frac{7992}{27} f(3) \\
& =296 f(3) .
\end{aligned}
$$

Therefore, because $f(3) \neq 0$, it follows that
$$
\frac{f(20)-f(2)}{f(3)}=296
$$

For INVESTIGATION
23.1 What is the value of
$$
\frac{f(46)-f(23)}{f(23)} ?
$$
23.2 Show that there are infinitely many positive integers $x, y$ and $z$ which satisfy the equation
$$
\frac{f(x)-f(y)}{f(z)}=7 \text {. }
$$
23.3 Show that for each positive integer $n$ the equation
$$
\frac{f(x)-f(y)}{f(z)}=n
$$
has either infinitely many solutions in which $x, y$ and $z$ are positive integers, or no such solutions.
23.4 Show that a function $g$ satisfies the equation $y^{3} g(x)=x^{3} g(y)$ for all real numbers $x$ and $y$ if, and only if, there is a constant $k$ such that $g(x)=k x^{3}$, for all real numbers $x$.