document.write( "Question 697462: What two consecutive integers have cubes that differ by 631? \n" ); document.write( "

Algebra.Com's Answer #430038 by Positive_EV(69) $\"\"$ $\"About$

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Call the smaller number x, so the larger integer is x + 1. The difference of their cubes is 631, so the equation is:
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\n" ); document.write( "\n" ); document.write( " $\"%28x%2B1%29%5E3+-+x%5E3+=+631\"$ , expanding: $\"%28x%2B1%29%5E3+=+x%5E3+%2B+3%2Ax%5E2+%2B+3%2Ax+%2B+1\"$ :
\n" ); document.write( " $\"x%5E3+%2B+3%2Ax%5E2+%2B+3%2Ax+%2B+1+-+x%5E3+=+631\"$
\n" ); document.write( " $\"3%2Ax%5E2+%2B+3%2Ax+%2B+1+=+631\"$ , now we have a quadratic equation that can be solved for x:
\n" ); document.write( " $\"3%2Ax%5E2+%2B+3%2Ax+-+630+=+0\"$ , divide both sides by 3 to simplify:
\n" ); document.write( " $\"x%5E2+%2B+x+-+210+=+0\"$ , this can be factored as
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\n" ); document.write( "\n" ); document.write( "(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:
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\n" ); document.write( "\n" ); document.write( " $\"15%5E3+-+14%5E3+=+3375+-+2744+=+631\"$ \n" ); document.write( "

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