document.write( "Question 985582: If x^2 + y^2 = 2 and xy = -1, then x^3 + y^3 = ? \n" ); document.write( "
Algebra.Com's Answer #606532 by ikleyn(52814)\"\" \"About 
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\n" ); document.write( "You have a system of two equations  (non-linear)  in two unknowns\r
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\n" ); document.write( "\n" ); document.write( "\"system%28x%5E2+%2B+y%5E2+=+2%2C%0D%0Axy+=+-1%29\".\r
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\n" ); document.write( "\n" ); document.write( "Express  \"y\"  from the second equation  \"y\" = \"-1%2Fx\"  and substitute into the first equation.  You will have \r
\n" ); document.write( "\n" ); document.write( "\"x%5E2\" + \"1%2Fx%5E2\" = \"2\".\r
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\n" ); document.write( "\n" ); document.write( "Introduce  \"z\" = \"x%5E2\".  You will have\r
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\n" ); document.write( "\n" ); document.write( "\"z\" + \"1%2Fz\" = \"2\",   \"z%5E2\" - \"2z\" + \"1\" = \"0\",   \"%28z-1%29%5E2\" = \"0\",   \"z\" = \"1\".\r
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\n" ); document.write( "\n" ); document.write( "Hence,  x = \"sqrt%28z%29\" = +/- 1.\r
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\n" ); document.write( "\n" ); document.write( "Correspondingly,  y = -/+ 1.\r
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\n" ); document.write( "\n" ); document.write( "Therefore,  the pairs  (x,y) = (1, -1)  and  (x,y) = (-1,1)  are the solutions.   There are no other solutions.\r
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\n" ); document.write( "\n" ); document.write( "It implies that   \"x%5E3\" + \"y%5E3\" = 0.\r
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