Math Help!!!!!!!!!!!!!!!!!

The product of three consecutive odd integers reduced by 23 is 99 less than the cube of the smallest number and 2. Compute the mean of the three integers.

x = 1st odd consecutive integer
x+2 = 2nd odd consecutive integer
x+4 = 3rd odd consecutive integer

x(x+2)(x+4) - 23 = (x+2)^3 - 99
x(x^2+6x+8) - 23 = (x+2)(x^2+4x+4) - 99
x^3+6x^2+8x-23 = x^3+4x^2+4x+2x^2+8x+8-99
68 = 4x
17 = x, so the odd consecutive integers are: 17, 19 and 21 and the mean is (17+19+21)/3 = 19

let the first integer be denoted by X. This means the other integers are (X+2) aintegers
The product of these integers reduced by 23 is:
(X)(X+2)(X+4)-23
Expanding this gives
X^3 +6X^2+8X-23
99 less than the cube of the smallest integer and 2 is:
(X+2)^3-99=X^3+6X^2+12X+8-99
Therefore
X^3+6X^2+8X-23=X^3+6X^2+12X+8-99
(X^3)-(X^3)+(6X^2)-(6X^2)+(8X)-(12X)=(…
-4X=-68
X=17
The three odd integers are 17,19 and 21
The mean of these numbers is imputed as:
(17+19+21)/3=(56)/3=18.67
I hope i've been of help 2 u

N*(N+2)*(N+3) - 23+99 = (N+2)^3
N^3+6N^2+8N -23+99 =N^3+6N^2+12N + 8
68 = 4N
N=17
Numbers are 17 , 19 , 21
Mean by inspection = 19
By calculation (17+19+21)/3 = 19

Are you sure of this wording? What is meant by 99 ess than smallest and 2?