Solution:

The conditions say $3b + 6$ is a perfect square and $2b + 7$ is a perfect cube. Since $2b + 7$ is odd and $b \lt 1000$ forces $2b + 7 \lt 2007,$ the cube must be one of $1,$ $27,$ $125,$ $343,$ $729,$ $1331,$ giving $b = -3,$ $10,$ $59,$ $168,$ $361,$ $662.$

The corresponding values of $3b + 6$ for the positive candidates are $36,$ $183,$ $510,$ $1089,$ $1992,$ and only $36 = 6^2$ (for $b = 10$) and $1089 = 33^2$ (for $b = 361$) are perfect squares. The requested sum is $10 + 361 = 371.$