document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209731>Question 1209731</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let r, s, and t be solutions of the equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E3+%2B+2x%5E2+-+5x+%2B+15+=+0 ALIGN=MIDDLE  ALT=\"x%5E3+%2B+2x%5E2+-+5x+%2B+15+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "Compute<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F%28r+-+2s+-+2t%29+%2B+1%2F%28s+-+2r+-+2t%29+%2B+1%2F%28t+-+2r+-+2s%29 ALIGN=MIDDLE  ALT=\"1%2F%28r+-+2s+-+2t%29+%2B+1%2F%28s+-+2r+-+2t%29+%2B+1%2F%28t+-+2r+-+2s%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850032 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</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=850032\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the given cubic equation be $x^3 + 2x^2 - 5x + 15 = 0$.<BR>\n" );
document.write( "Let $r, s, t$ be the roots of the equation.<BR>\n" );
document.write( "By Vieta's formulas, we have:<BR>\n" );
document.write( "\begin{align*} \label{eq:1} r + s + t &= -2 \\ rs + st + tr &= -5 \\ rst &= -15\end{align*} <BR>\n" );
document.write( "We want to compute<BR>\n" );
document.write( "$$ S = \frac{1}{r - 2s - 2t} + \frac{1}{s - 2r - 2t} + \frac{1}{t - 2r - 2s} $$<BR>\n" );
document.write( "We can rewrite the denominators as follows:<BR>\n" );
document.write( "\begin{align*} r - 2s - 2t &= r - 2(s+t) = r - 2(-2-r) = r + 4 + 2r = 3r + 4 \\ s - 2r - 2t &= s - 2(r+t) = s - 2(-2-s) = s + 4 + 2s = 3s + 4 \\ t - 2r - 2s &= t - 2(r+s) = t - 2(-2-t) = t + 4 + 2t = 3t + 4\end{align*} <BR>\n" );
document.write( "Then,<BR>\n" );
document.write( "$$ S = \frac{1}{3r+4} + \frac{1}{3s+4} + \frac{1}{3t+4} $$<BR>\n" );
document.write( "Let $y = 3x+4$, so $x = \frac{y-4}{3}$. Substituting this into the cubic equation, we get<BR>\n" );
document.write( "$$ \left(\frac{y-4}{3}\right)^3 + 2\left(\frac{y-4}{3}\right)^2 - 5\left(\frac{y-4}{3}\right) + 15 = 0 $$<BR>\n" );
document.write( "$$ (y-4)^3 + 6(y-4)^2 - 45(y-4) + 405 = 0 $$<BR>\n" );
document.write( "$$ y^3 - 12y^2 + 48y - 64 + 6(y^2 - 8y + 16) - 45y + 180 + 405 = 0 $$<BR>\n" );
document.write( "$$ y^3 - 12y^2 + 48y - 64 + 6y^2 - 48y + 96 - 45y + 180 + 405 = 0 $$<BR>\n" );
document.write( "$$ y^3 - 6y^2 - 45y + 622 = 0 $$<BR>\n" );
document.write( "The roots of this equation are $3r+4$, $3s+4$, and $3t+4$. Let $u = 3r+4$, $v = 3s+4$, $w = 3t+4$.<BR>\n" );
document.write( "Then<BR>\n" );
document.write( "$$ \frac{1}{u} + \frac{1}{v} + \frac{1}{w} = \frac{uv+vw+uw}{uvw} $$<BR>\n" );
document.write( "From the equation $y^3 - 6y^2 - 45y + 622 = 0$, we have<BR>\n" );
document.write( "\begin{align*} u+v+w &= 6 \\ uv+vw+uw &= -45 \\ uvw &= -622 \end{align*} <BR>\n" );
document.write( "Therefore,<BR>\n" );
document.write( "$$ S = \frac{1}{3r+4} + \frac{1}{3s+4} + \frac{1}{3t+4} = \frac{uv+vw+uw}{uvw} = \frac{-45}{-622} = \frac{45}{622} $$\r<BR>\n" );
document.write( "\n" );
document.write( "Final Answer: The final answer is $\boxed{45/622}$<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );