document.write( "Question 896003: find three consecutive integers whose product is 56 larger than the cube of the smallest integer \n" ); document.write( "
Algebra.Com's Answer #805820 by CubeyThePenguin(3113)![]() ![]() You can put this solution on YOUR website! consecutive integers: (x-1), x, (x+1)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(x-1)(x)(x+1) = 56 + (x-1)^3 \n" ); document.write( "x^3 - x = x^3 - 3x^2 + 3x + 55 \n" ); document.write( "0 = -3x^2 + 4x + 55 \n" ); document.write( "0 = 3x^2 - 4x - 55 \n" ); document.write( "0 = (x - 5)(3x + 11) \n" ); document.write( "x = 5 or x = -11/3\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x has to be an integer, so the integers are 4, 5, and 6. \n" ); document.write( " |