document.write( "Question 1105564: (x+y)(x^2+y^2)=40
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Algebra.Com's Answer #720418 by ikleyn(52835) $\"\"$ $\"About$

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document.write( "(x+y)(x^2+y^2)=40    (1)\r\n" );
document.write( "(x-y)(x^2-y^2)=16    (2)\r\n" );
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document.write( "=================>\r\n" );
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document.write( "x^3 + xy^2 + yx^2 + y^3 = 40   (1')\r\n" );
document.write( "x^3 - xy^2 - yx^2 + y^3 = 16   (2')\r\n" );
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document.write( "Add (1') and {2') (both sides). You will get\r\n" );
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document.write( "2(x^3 + y^3) = 56,     or  x^3 + y^3 = 28.     (3)\r\n" );
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document.write( "Subtract (2') from (1'). You will get\r\n" );
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document.write( "2(xy^2 + yx^2) = 24,   or  xy^2 + x^2y = 12.   (4)\r\n" );
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document.write( "Then\r\n" );
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document.write( "x^3 + 3x^2y + 3xy^2 + y^3 = 28 + 3*12 = 64,   or\r\n" );
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document.write( "(x+y)^3 = 64,   which implies   \r\n" );
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document.write( "x + y = 4       (5)\r\n" );
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document.write( "Next, from (4) you have\r\n" );
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document.write( "xy*(x+y) = 12,  and,  replacing here x+y by 4 (due to (5)), you get\r\n" );
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document.write( "xy*4 = 12,   or\r\n" );
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document.write( "xy =  = 3.    (6)\r\n" );
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document.write( "From (5) and (6) you can easily get the \r\n" );
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document.write( "Answer.  (x,y) = (1,3)   or  (x,y) = (3,1).\r\n" );
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