document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=548600>Question 548600</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The solutions of x^3 – 2x^2 + 3x + 4 = 0 are a, b, and c. what is the numerical value of (a + b)(a + c)(b + c) ?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #358053 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=358053\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> By Vieta's formulas, a+b+c = 2. Hence we are evaluating (2-c)(2-b)(2-a). We could expand this entire expression (since we are guaranteed to obtain useful cyclic sums:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (2-c)(2-b)(2-a) = 8 - 4a - 4b - 4c + 2ab + 2ac + 2bc - abc\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE = 8 - 4(a+b+c) + 2(ab + ac + bc) - abc\">\r<BR>\n" );
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document.write( "However, we have a+b+c = 2, ab+ac+bc = 3, abc = 4 (you can verify these by expanding (x-a)(x-b)(x-c) and equating like terms). Hence, the expression we want is equal to\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 8 - 4(2) + 2(3) - 4 = 2\"> </SPAN>\n" );
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