When a bactericide is added to a nutrient broth in which bacteria are growing, the bacterium population continues to grow for a while, but then stops growing and begins to decline. The size of the population at time $t$ (hours) is $b=6^{7}+6^{5} t-6^{4} t^{2}$. Find the growth rates at $t=0$ hours, $t=3$ hours, and $t=6$ hours.

Step 1 ：The size of the bacteria population at time $t$ (hours) is given by the function $b=6^{7}+6^{5} t-6^{4} t^{2}$.

Step 2 ：To find the growth rate of the bacteria population, we need to take the derivative of the population function with respect to time. This will give us a new function that describes the rate of change of the population at any given time.

Step 3 ：The derivative of the population function is $db/dt = 7776 - 2592t$.

Step 4 ：We can now evaluate this function at the given times to find the growth rates.

Step 5 ：At $t=0$ hours, the growth rate is $7776$.

Step 6 ：At $t=3$ hours, the growth rate is $0$.

Step 7 ：At $t=6$ hours, the growth rate is $-7776$.

Step 8 ：Final Answer: The growth rates of the bacteria population at $t=0$ hours, $t=3$ hours, and $t=6$ hours are \(\boxed{7776}\), \(\boxed{0}\), and \(\boxed{-7776}\) respectively.