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Question 894661: Can you find three consecutive integers whose product is 5 greater than the cube of the middle integer?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the number is x = -6
-6*-5*-4 = -120
(-5)^3 + 5 = -125 + 5 = -120
the number checks out ok.
the process is as follows:
the 3 numbers are x, x+1, x+2
(x+1)^3 + 5 = x^3 + 3x^2 + 3x + 1 + 5 which is equal to:
x^3 + 3x^2 + 3x + 6
x*(x+1)*(x+2) is equal to x^3 + 3x^2 + 2x
you get:
x^3 + 3x^2 + 2x = x^3 + 3x^3 + 3x + 6
subtract (x^3 + 3x^2 + 2x) from both sides of this equation to get:
0 = x + 6
solve for x to get x = -6
the rest is confirmation that was done up top.
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