Algebra.Com's Answer #705084 by ikleyn(52781)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "If x^3 - y^3 = 485 and x-y = 5, find the numerical value of x + y.
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document.write( "Recall factoring (short cut) x^3 - y^3 = (x-y)*(x*2 + xy + y^2).\r\n" );
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document.write( "Then you have 485 = x^3 - y^3 = (x-y)*(x^2 + xy + y^2) = (replace x-y by 5) = 5*(x^2 + xy + y^2), \r\n" );
document.write( "which implies (after canceling the factor 5)\r\n" );
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document.write( " = (x^2 + xy + y^2).\r\n" );
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document.write( "So, you have this equation \r\n" );
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document.write( "x^2 + xy + y^2 = 97. (1)\r\n" );
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document.write( "From x-y = 5 express y = x-5 and substitute it for y into equation (1). You will get\r\n" );
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document.write( "x^2 + x*(x-5) + (x-5)^2 = 97.\r\n" );
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document.write( "Simplify it step by step and then solve for x:\r\n" );
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document.write( "x^2 + x^2 - 5x + x^2 - 10x + 25 = 97,\r\n" );
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document.write( "3x^2 - 15x - 72 = 0,\r\n" );
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document.write( "x^2 - 5x - 24 = 0. (2)\r\n" );
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document.write( "Factor left side to get an equivalent equation\r\n" );
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document.write( "(x-8)*(x+3) = 0.\r\n" );
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document.write( "The roots are x = 8 and x = -3.\r\n" );
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document.write( "Answer. One solution is x= 8, y= 3. Then x + y = 11.\r\n" );
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document.write( " The other solution is x= -3, y= -8. Then x + y = -11.\r\n" );
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document.write( "Notice. What I actually did in the course of solution, was reducing the given equation of the degree 3 to the quadratic equation
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document.write( " using given information.\r
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document.write( "Solved.\r
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