1.       The squares of three consecutive odd numbers total eight more than nine times the sum of the three numbers. What are the three numbers?

Let n be the first odd number. Then we can write the next two as (n+2) and (n+4). eg. if first odd is 5 then next two is 5+2 = 7 and 5+4 = 9.

It is given that the squares of three consecutive odd numbers total eight more than nine times the sum of the three numbers.

`n^2+n^2+4n+4+n^2+8n+16 = 9(3n+6)+8`

`3n^2+12n+20 = 27n+54+8`

`3n^2-15n-42 = 0`

`n = (-(-15)+-sqrt((-15^2)-4xx3xx(-42)))/(2xx3)`

n = 7 and n = -2.

So we get n = 7 as the answer of the first positive number. Then next two will be 7+2 = 9 and 7+4 = 11.

So the three numbers are  7,9 and 11.