document.write( "Question 894661: Can you find three consecutive integers whose product is 5 greater than the cube of the middle integer?
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Algebra.Com's Answer #542219 by Theo(13342)![]() ![]() You can put this solution on YOUR website! the number is x = -6\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-6*-5*-4 = -120\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(-5)^3 + 5 = -125 + 5 = -120\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the number checks out ok.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the process is as follows:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the 3 numbers are x, x+1, x+2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(x+1)^3 + 5 = x^3 + 3x^2 + 3x + 1 + 5 which is equal to:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x^3 + 3x^2 + 3x + 6\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x*(x+1)*(x+2) is equal to x^3 + 3x^2 + 2x\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you get:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x^3 + 3x^2 + 2x = x^3 + 3x^3 + 3x + 6\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "subtract (x^3 + 3x^2 + 2x) from both sides of this equation to get:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "0 = x + 6\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for x to get x = -6\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the rest is confirmation that was done up top.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |