Carefully Divide

$\dfrac{1+2+4+8+16+32+63+128+256}{1+2+4}=\dfrac{2^9-1}{2^3-1}$

it is of the form $\dfrac{a^3-b^3}{a-b}$ ( $a=2^3,b=1$ )

$\begin{aligned}\dfrac{a^3-b^3}{a-b}&=a^2+ab+b^2\\&=2^6+2^3+1=\boxed{1+8+64}\end{aligned}$