jmc

If we are working in base $b$, then we have $(4b+4)(5b+5)-3b^3-5b^2-6=0$. $$\begin{array}{rl}0& =(4b+4)(5b+5)-3b^3-5b^2-6\\ & =20(b+1{)}^2-3b^3-5b^2-6\\ & =20b^2+40b+20-3b^3-5b^2-6\\ & =-3b^3+15b^2+40b+14\end{array}$$Therefore, we must solve the cubic $3b^3-15b^2-40b-14=0$. By the Rational Root Theorem, the only possible positive integer solutions to this equation are 1, 2, 7, and 14. 1 and 2 are invalid bases since the digit 6 is used, so we first try $b=7$. It turns out that $b=7$ is a solution to this cubic. If we divide by $b-7$, we get the quadratic $3b^2+6b+2$, which has no integral solutions. Therefore, in base $\boxed{7}$, we have $44\times 55=3506$.