Only 100 of base b, what's b?

Here's an approach to find this

Note that  in base 10

999 - 100 + 1   =  900 three digit integers

So.....we really have

[ 9*10^2 + 9 * 10 + (10 - 1)]  - 10^2 + 1  = 900

So...considering  base 10, we have

 [ (b - 1) * b^2 + (b - 1) * b  + (b - 1) ]  - b^2 + 1  = 900

But....we are looking for the base that only produces 100  three digit integers

So......We can solve this to find the base, b

[ (b - 1) * b^2 + ( b - 1) * b + ( b - 1) ] - b^2 + 1 = 100      simplify

b^3 - b^2 + b^2 - b  + b - 1 - b^2 + 1 = 100

b^3 - b^2  = 100

And note that this is true for b = 5

Bruh, seriously you're asking on this website for AOPS homework answers. Bruh