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+8x23 = x^3+4x^2+4x+2x^2+8x+899
68 = 4x
17 = x, so the odd consecutive integers are: 17, 19 and 21 and the mean is (17+19+21)/3 = 19
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+8x23 = x^3+4x^2+4x+2x^2+8x+899
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+8X23
99 less than the cube of the smallest integer and 2 is:
(X+2)^399=X^3+6X^2+12X+899
Therefore
X^3+6X^2+8X23=X^3+6X^2+12X+899
(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
The product of these integers reduced by 23 is:
(X)(X+2)(X+4)23
Expanding this gives
X^3 +6X^2+8X23
99 less than the cube of the smallest integer and 2 is:
(X+2)^399=X^3+6X^2+12X+899
Therefore
X^3+6X^2+8X23=X^3+6X^2+12X+899
(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
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?