document.write( "Question 1012813: If 1/(a-(-w))-(-1)/(b-(-w))-(-1)/(c -(-w))-(-1)/(d-(-w))=2w^2 and 1/(a-(-w^2))-(-1)/(b-(-w^2))-(-1)/(c-(-w^2))-(-1)/(d-(-w^2))=2w then prove that 1/(a-(-1))-(-1)/(b-(-1))-(-1)/(c-(-1))-(-1)/(d-(-1))=2
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Algebra.Com's Answer #628915 by ikleyn(52788)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Now your formulas look much better.\r\n" );
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document.write( "Thank you for following my instructions.\r\n" );
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document.write( "Now tell us please what is \"w\" in these formulas.\r\n" );
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document.write( "OK, very good. \r\n" );
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document.write( "So,  w = \"sqrt%281%29\" = \"%28-1%2F2%29+%2B+i%2A%28sqrt%283%29%2F2%29\"  is this complex number.\r\n" );
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document.write( "Let me rewrite what is given and what you want to prove in an equivalent, but simpler form:\r\n" );
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If \r\n" ); document.write( "\r\n" ); document.write( "\"1%2F%28a%2Bw%29+%2B+1%2F%28b%2Bw%29+%2B+1%2F%28c%2Bw%29+%2B+1%2F%28d%2Bw%29\" = \"2w%5E2\" (1) and \r\n" ); document.write( "\r\n" ); document.write( "\"1%2F%28a%2Bw%5E2%29+%2B+1%2F%28b%2Bw%5E2%29+%2B+1%2F%28c%2Bw%5E2%29+%2B+1%2F%28d%2Bw%5E2%29\" = \"2w\" (2) then prove that \r\n" ); document.write( "\r\n" ); document.write( "\"1%2F%28a%2B1%29+%2B+1%2F%28b%2B1%29+%2B+1%2F%28c%2B1%29+%2B+1%2F%28d%2B1%29\" = \"2\".
\r\n" ); document.write( "\r\n" ); document.write( "Let us add the formulas (1) and (2) term-by-term. I will start with the pair of addends with the symbol \"a\", and will simplify it step by step:\r\n" ); document.write( "\r\n" ); document.write( "\"1%2F%28a%2Bw%29\" + \"1%2F%28a%2Bw%5E2%29\" = (write with the common denominator and then add the numerators. You will get)\r\n" ); document.write( "\r\n" ); document.write( "= \"%28%28a%2Bw%5E2%29+%2B+%28a%2Bw%29%29%2F%28%28a%2Bw%29%2A%28a%2Bw%5E2%29%29\" = \"%282a+%2B+w+%2B+w%5E2%29%2F%28a%5E2+%2Baw+%2B+aw%5E2+%2B+w%5E3%29\" = (now recall that \"w%5E3\" = 1. So, replace \"w%5E3\" by 1 in the last formula. You will get)\r\n" ); document.write( "\r\n" ); document.write( "= \"%282a+%2B+w+%2B+w%5E2%29%2F%281+%2B+a%5E2+%2B+a%28w+%2B+w%5E2%29%29\" = (now use the fact that for your special number \"w\" there is the identity \"w+%2B+w%5E2\" = -1. \r\n" ); document.write( " So, substitute it into the last formula. You will get)\r\n" ); document.write( "= \"%28-1%2B2a%29%2F%281+-+a+%2B+a%5E2%29\".\r\n" ); document.write( "\r\n" ); document.write( "OK. It looks like I can not simplify it further.\r\n" ); document.write( "\r\n" ); document.write( "You can make this trick with every other pair of addends containing the symbols \"b\", \"c\" and \"d\". \r\n" ); document.write( "In this way you will obtain similar formulas.\r\n" ); document.write( "\r\n" ); document.write( "Now, when you add right sides of these long formulas, you will get \"2w%5E2+%2B+2w%7D%7D%2C+which+is+%7B%7B%7B2%2A%28w+%2B+w%5E2%29\". It equals -2 based on what was noted above.\r\n" ); document.write( "\r\n" ); document.write( "Finally, after adding these long formulas and simplifications, you get\r\n" ); document.write( "\r\n" ); document.write( "\"%281+-+2a%29%2F%281+-+a+%2B+a%5E2%29\" + \"%281+-+2b%29%2F%281+-+b+%2B+b%5E2%29\" + \"%281+-+2c%29%2F%281+-+c+%2B+c%5E2%29\" + \"%281+-+2d%29%2F%281+-+d+%2B+d%5E2%29\" = 2.\r\n" ); document.write( "\r\n" ); document.write( "It is not exactly what you requested.\r\n" ); document.write( "But it is what our mother-nature allows us to do.\r\n" ); document.write( "

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