SOLUTION: What two consecutive integers have cubes that differ by 631?

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Question 697462: What two consecutive integers have cubes that differ by 631?
Found 2 solutions by Alan3354, Positive_EV:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
What two consecutive integers have cubes that differ by 631?
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5 & 6

Answer by Positive_EV(69) About Me  (Show Source):
You can put this solution on YOUR website!
Call the smaller number x, so the larger integer is x + 1. The difference of their cubes is 631, so the equation is:

%28x%2B1%29%5E3+-+x%5E3+=+631, expanding: %28x%2B1%29%5E3+=+x%5E3+%2B+3%2Ax%5E2+%2B+3%2Ax+%2B+1:
x%5E3+%2B+3%2Ax%5E2+%2B+3%2Ax+%2B+1+-+x%5E3+=+631
3%2Ax%5E2+%2B+3%2Ax+%2B+1+=+631, now we have a quadratic equation that can be solved for x:
3%2Ax%5E2+%2B+3%2Ax+-+630+=+0, divide both sides by 3 to simplify:
x%5E2+%2B+x+-+210+=+0, this can be factored as

(x+15)(x-14) = 0. So, x can be 14 or -15. Since the problem doesn't specify if the integers are positive or negative, both solutions are fine: 14 and 15 or -14 and -15 are both acceptable. Check:

15%5E3+-+14%5E3+=+3375+-+2744+=+631