|
Question 813470: the difference between the cubes of the two larger of the three consecutive integers is 66 more than the difference between the cubes of two smaller integers. What is the median integer?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! let x = the median integer; then the other two integers are (x-1), x+1)
:
the difference between the cubes of the two larger of the three consecutive integers is 66 more than the difference between the cubes of two smaller integers.
(x+1)^3 - x^3 = x^3 - (x-1)^3 + 66
:
Cube (x+1) and (x-1)
x^3+3x^2+3x+1 - x^3 = x^3 - (x^3-3x^2+3x-1)+ 66
:
removing the brackets changes the signs
x^3 + 3x^2 + 3x + 1 - x^3 = x^3 - x^3 + 3x^2 - 3x + 1 + 66
:
x^3's cancel (fortunately)
3x^2 + 3x + 1 = 3x^2 - 3x + 67
:
Combine like terms on the left, leave 66 where it is
3x^2 - 3x^2 + 3x + 3x + 1 - 1 = 66
6x = 66
x = 11 is the median integer
:
:
See if that checks out
12^3 - 11^3 = 11^3 - 10^3 + 66
1728 - 1331 = 1331 - 1000 + 66
397 = 331 + 66
|
|
|
| |
