SOLUTION: The product of three consecutive odd integers is reduced by 23. That result is 99 less than the cube of the sum of the number 2 and the smallest of the three consecutive odd intege

Question 737355: The product of three consecutive odd integers is reduced by 23. That result is 99 less than the cube of the sum of the number 2 and the smallest of the three consecutive odd integers.
Determining the mean of the three consecutive odd integers is your task.
Answer by sachi(548) (Show Source): You can put this solution on YOUR website!
let the three consecutive odd integers are a-2,a,a+2
The product of three consecutive odd integers is reduced by 23. That result is 99 less than the cube of the sum of the number 2 and the smallest of the three consecutive odd integers.
so (a-2)a(a+2)-23=-99+[2+(a-2)]^3=-99+a^3
or a^3-4a-a^3=-99+23=-76
or -4a=-76
or a=19
so the nos are 17,19,21
the mean of the three consecutive odd integers =(17+19+21)/3=19
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