document.write( "Question 1145006: If a, b, c are the zeroes of the polynomials p(x) = x³ + 4, then find the value of
\n" ); document.write( "k = - 3[(1/(a² - a + 1) + (1/(b² - b + 1)] + c \n" ); document.write( "
Algebra.Com's Answer #766206 by ikleyn(52781)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "            Surely and certainly,  this problem is advanced and is intended for advanced students. \r
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document.write( "(1)  \"1%2F%28a%5E2+-+a+%2B+1%29\" = \"%28a%2B1%29%2F%28a%5E3%2B1%29\".\r\n" );
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document.write( "     It follows from the standard identity  \"a%5E3%2B1\" = \"%28a%2B1%29%2A%28a%5E2-a%2B1%29\".\r\n" );
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document.write( "     Further, since \"a\" is the root of the equation  \"x%5E3%2B4\" = 0,  we have  \r\n" );
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document.write( "         \"a%5E3%2B4\" = 0;  hence,   \"a%5E3%2B1\" = \"%28a%5E3%2B4%29-3\" = \"0-3\" = -3.   \r\n" );
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document.write( "     Therefore,  \"1%2F%28a%5E2+-+a+%2B+1%29\" = \"%28a%2B1%29%2F%28a%5E3%2B1%29\" = \"-%28a%2B1%29%2F3\".    (1)\r\n" );
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document.write( "(2)  Similarly,  \"1%2F%28b%5E2+-+b+%2B+1%29\" = \"-%28b%2B1%29%2F3\".             (2)\r\n" );
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document.write( "(3)  Therefore,\r\n" );
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document.write( "         k = - 3[(1/(a² - a + 1) + (1/(b² - b + 1)] + c  = -3*( \"-%28a%2B1%29%2F3+-+%28b%2B1%29%2F3\" ) + c = (a+1) + (b+1) + c = (a + b + c) + 2.\r\n" );
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document.write( "      The sum (a + b + c) is the coefficient of the given polynomial  p(x) = x^3 + 4  at \"x\"; so, it is 0 (zero, ZERO) :\r\n" );
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document.write( "          a + b + c = 0.\r\n" );
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document.write( "      Therefore,  k = 0 + 2 = 2.     ANSWER\r\n" );
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document.write( "ANSWER.  k = 2.\r\n" );
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