SOLUTION: What two consecutive integers have cubes that differ by 217?

Question 155016: What two consecutive integers have cubes that differ by 217?
Found 2 solutions by checkley77, Earlsdon:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
(x+1)^3-x^3=217
x^3+3x^2+3x+1-x^3=217
3x^2+3x-216=0
(3x-24)(x+9)=0
3x-24=0
3x=24
x=24/3
x=8 for the amallernteger.
8+1=9 for the larger integer.
proof;
9^3-8^3=217
729-512=217
217=217

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let the first integer be x, the next consecutive integer is (x+1).
Expand this.
Simplify.
Subtract 217 from both sides.
Factor the trinomial.

or
The two integers are 8 and 9
Check:
=
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