document.write( "Question 1132448: The product of three consecutive integers is 21 more than the cube of the smallest integer. The smallest integer is
\n" ); document.write( "A) -3
\n" ); document.write( "B) -4
\n" ); document.write( "C) -5
\n" ); document.write( "D) -6
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Algebra.Com's Answer #749473 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "Let the three integers be x, x+1, and x+2. Then

\n" ); document.write( "the product of the three integers is x(x+1)(x+2) = x(x^2+3x+2) = x^3+3x^2+2x; and
\n" ); document.write( "the cube of the smallest integer is x^3.

\n" ); document.write( "The problem says the product of the 3 is 21 more than the cube of the smallest:

\n" ); document.write( "\"x%5E3%2B3x%5E2%2B2x+=+x%5E3%2B21\"

\n" ); document.write( "The x^3 terms cancel, leaving a quadratic equation which is easily solved by any of a number of different methods.

\n" ); document.write( "Of course, if an algebraic solution is not required, the answer is easily found by trying the given answer choices....
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