SOLUTION: find 3 consecutive integers such that the product of all 3, decreased by the cube of the first, is 33

SOLUTION: find 3 consecutive integers such that the product of all 3, decreased by the cube of the first, is 33

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Question 390658: find 3 consecutive integers such that the product of all 3, decreased by the cube of the first, is 33
Answer by checkley79(3341) (Show Source): You can put this solution on YOUR website!
Let x, x+1, x+2 be the 3 integers.
x(x+1)(x+2)-x^3=33
(x^2+x)(x+2)-x^3=33
x^3+x^2+2x^2+2x-3x^3=33
3x^2+2x=33
3x^2+2x-33=0
(3x+11)(x-3)=0
x-3=0
x=3 ans.
Proof:
3(3+1)(3+2)-3^3=33
3*4*5-27=33
60-27=33
33=33
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