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Find three consecutive integers whose products is 85 larger than the cube of the smallest integer

User Plaes
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1 Answer

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Let the numbers be x, x + 1 and x + 2.

From the information given:-

x(x + 1)(x + 2) = x^3 + 85

Solving for x:-

x(x^2 + 3x + 2) = x^3 + 85

x^3 + 3x^2 + 2x = x^3 + 85

3x^2 + 2x - 85 = 0

3x^2 - 15x + 17x - 85 = 0

3x(x - 5) + 17(x - 5) = 0

(3x + 17)(x - 5) = 0

x = -17/3 , 5

We take x = 5 because x is an integer.

Answer:- the 3 integers are 5, 6 and 7.



User Arbulgazar
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