Suppose two cubic polynomials f (x) and g(x) satisfy the following: f (2) = g(4); f (4) = g(8); f (8) = g(16); f (16) = g(32) + 64. Suppose two cubic polynomials $f (x)$ and $g(x)$ satisfy the following: If $f(2) = g(4)$; $f (4) = g(8)$;
$f (8) = g(16)$; $f (16) = g(32) + 64$. What is the value of $g(128) − f (64)$?
The first thing that comes to mind is that the polynomials must be in the form $ax^3 + b$ because 4 restrictions are given, but this question was given in an oral competition and I don't think my method is appropriate.

Let $F(x)=g(2x)-f(x) $. We have $F(2)=0,F(4)=0,F(8)=0$. So
\begin{eqnarray*}
F(x) = \lambda (x-2)(x-4)(x-8).
\end{eqnarray*}
Using $F(16)=-64$ gives
\begin{eqnarray*}
F(x) = -\frac{(x-2)(x-4)(x-8)}{3 \times 7} .
\end{eqnarray*}
So
\begin{eqnarray*}
F(64) = -\frac{62 \times 60 \times 56}{3 \times 7} =- 8 \times 20 \times 62 = \color{red}{-9920}.
\end{eqnarray*}
