SOLUTION: Can you find three consecutive integers whose product is 5 greater than the cube of the middle integer?

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.

RELATED QUESTIONS
find three consecutive integers whose product is 85 larger than the cube of the smallest... (answered by stanbon)

Find three consecutive integers whose product is 161 larger than the cube of the smallest (answered by ewatrrr,MathTherapy)

find three consecutive integers whose product is 56 larger than the cube of the smallest... (answered by CubeyThePenguin)

The product of three consecutive integers is 22 less than the cube of the middle integer. (answered by citychic3)

Find three consecutive integers whose product is 208 larger that the cube of the smallest (answered by CubeyThePenguin)

Find three consecutive integers whos product is 120 larger than the cube of the smallest... (answered by solver91311)

The product of three consecutive integers is 21 more than the cube of the smallest... (answered by DrBeeee,algebrahouse.com)

what is the product of three consecutive odd integers if the middle integer is p-5 (answered by Edwin McCravy)