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.
 A: Consider the polynomial
$$
h(x) = f(x)-g(2x)
$$
It is a cubic polynomial, and we have
$$
h(2) = h(4) = h(8) = 0\\
h(16) = 64
$$
We are asked to find $-h(64)$.
The three roots of $h$ means that it is of the form
$$
h(x) = a(x-2)(x-4)(x-8)
$$
And the final condition implies that
$$
a = \frac{64}{(16-2)(16-4)(16-8)} = \frac{1}{21}
$$
Now we can just calculate $-h(64)$ immediately:
$$
-h(64) = -\frac1{21}(64-2)(64-4)(64-8) = -\frac1{21}\cdot 62\cdot 60\cdot 56 = -9920
$$
A: Define $h(x)=g(2x)$.  Then we are given that $f(x)$ and $h(x)$ agree at $2,4,8$ and $f(16)=h(16)+64$, so $h(x)-f(x)=0$ at $2,4,8$ and $64$ at $16$.  We need to find a cubic that goes through these four points.  
Knowing the roots we know $h(x)-f(x)=a(x-2)(x-4)(x-8)$, so we write 
$$64=a(16-2)(16-4)(16-8)=1344a\\a=\frac{1}{21}$$
Then $$f(64)-g(128)=f(64)-h(64)\\
=-\frac{1}{21}(64-2)(64-4)(64-8)\\
=-9920$$
A: Let
$$g(2x)=f(x)+k(x-2)(x-4)(x-8)$$
$$Put\, x=16$$
$$g(32)=f(16)+k(14)(12)(8)=f(16)-64$$
$$K=-1/21$$
Hence 
$$g(128)-f(64)=-(1/21)(62)(60)(56)=-9920$$
A: 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*}
