If we are working in base $ b $, then we have $ (4b+4)(5b+5) - 3b^3 - 5b^2 - 6 = 0 $. \begin{align*} 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{align*}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 $.