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

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Question 155016: What two consecutive integers have cubes that differ by 217?
Found 2 solutions by checkley77, Earlsdon:
Answer by checkley77(12844) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
Let the first integer be x, the next consecutive integer is (x+1).
%28x%2B1%29%5E3+-+x%5E3+=+217 Expand this.
%28x%5E3%2B3x%5E2%2B3x%2B1%29-x%5E3+=+217 Simplify.
3x%5E2%2B3x%2B1+=+217 Subtract 217 from both sides.
3x%5E2%2B3x-216+=+0 Factor the trinomial.
3%28x%5E2%2Bx-72%29+=+0
x%5E2%2Bx-72+=+0
%28x-8%29%28x%2B9%29+=+0
x+=+8 or x+=+-9
The two integers are 8 and 9
Check:
9%5E3-8%5E3+=+729-512=217