document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209738>Question 1209738</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let r, s, and t be solutions of the equation 3x^3 - 4x^2 - 2x + 12 = 0.  Compute<BR>\n" );
document.write( "\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2}. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850046 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850046\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this problem:\r<BR>\n" );
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document.write( "1. **Vieta's Formulas:**<BR>\n" );
document.write( "   For a cubic equation of the form ax³ + bx² + cx + d = 0, Vieta's formulas relate the coefficients to sums and products of the roots. In our case, 3x³ - 4x² - 2x + 12 = 0, so a = 3, b = -4, c = -2, and d = 12. The formulas we'll need are:\r<BR>\n" );
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document.write( "   * rst = -d/a = -12/3 = -4<BR>\n" );
document.write( "   * rs + rt + st = c/a = -2/3\r<BR>\n" );
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document.write( "2. **Rewrite the Expression:**<BR>\n" );
document.write( "   We want to compute:\r<BR>\n" );
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document.write( "   (rs/t²) + (rt/s²) + (st/r²) = (r³s³ + r³t³ + s³t³) / (r²s²t²)\r<BR>\n" );
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document.write( "3. **Simplify using Vieta's Formulas:**<BR>\n" );
document.write( "   We know that r²s²t² = (rst)² = (-4)² = 16.  So, we need to find r³s³ + r³t³ + s³t³.\r<BR>\n" );
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document.write( "4. **Key Identity:**<BR>\n" );
document.write( "   Recall the identity:  A³ + B³ + C³ - 3ABC = (A + B + C)(A² + B² + C² - AB - BC - CA)<BR>\n" );
document.write( "   Let A = rs, B = rt, and C = st.  Then:\r<BR>\n" );
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document.write( "   (rs)³ + (rt)³ + (st)³ - 3(rst)² = (rs + rt + st)[(rs)² + (rt)² + (st)² - rs*rt - rs*st - rt*st]\r<BR>\n" );
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document.write( "   We have rs + rt + st = -2/3 and rst = -4.  Substituting these values:\r<BR>\n" );
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document.write( "   r³s³ + r³t³ + s³t³ - 3(-4)² = (-2/3)[(rs)² + (rt)² + (st)² - r²st - rs²t - rst²]<BR>\n" );
document.write( "   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - rt(rs + st + rt)]<BR>\n" );
document.write( "   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - (-4)(-2/3)]<BR>\n" );
document.write( "   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - 8/3]\r<BR>\n" );
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document.write( "   We also know that (rs + rt + st)² = (rs)² + (rt)² + (st)² + 2rst(r + s + t).<BR>\n" );
document.write( "   (-2/3)² = (rs)² + (rt)² + (st)² + 2(-4)(4/3)<BR>\n" );
document.write( "   4/9 = (rs)² + (rt)² + (st)² - 32/3<BR>\n" );
document.write( "   (rs)² + (rt)² + (st)² = 4/9 + 96/9 = 100/9\r<BR>\n" );
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document.write( "   Now plug this back into the equation:<BR>\n" );
document.write( "   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[100/9 - 8/3]<BR>\n" );
document.write( "   r³s³ + r³t³ + s³t³ - 48 = (-2/3)[100/9 - 24/9]<BR>\n" );
document.write( "   r³s³ + r³t³ + s³t³ - 48 = (-2/3)(76/9)<BR>\n" );
document.write( "   r³s³ + r³t³ + s³t³ = 48 - 152/27<BR>\n" );
document.write( "   r³s³ + r³t³ + s³t³ = (1296 - 152)/27 = 1144/27\r<BR>\n" );
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document.write( "5. **Final Calculation:**\r<BR>\n" );
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document.write( "   (r³s³ + r³t³ + s³t³) / (r²s²t²) = (1144/27) / 16 = 1144 / (27 * 16) = 1144 / 432 = 143/54\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{143/54}$<BR>\n" );
document.write( " </SPAN>\n" );
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