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Let $r, s, t$ be the roots of the cubic equation $x^3 - 4x^2 - 7x + 12 = 0$.
\n" ); document.write( "By Vieta's formulas, we have:
\n" ); document.write( "\begin{align*} r+s+t &= 4 \\ rs+st+tr &= -7 \\ rst &= -12 \end{align*}
\n" ); document.write( "We want to compute
\n" ); document.write( "$$ S = \frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2} = \frac{r^3s^3 + r^3t^3 + s^3t^3}{r^2s^2t^2} $$
\n" ); document.write( "We know that $rst = -12$, so $r^2s^2t^2 = (rst)^2 = (-12)^2 = 144$.
\n" ); document.write( "Let $A = rs$, $B = st$, and $C = tr$. Then we want to compute
\n" ); document.write( "$$ \frac{A^3 + B^3 + C^3}{144} $$
\n" ); document.write( "We have $A+B+C = rs+st+tr = -7$.
\n" ); document.write( "Also, $ABC = (rs)(st)(tr) = (rst)^2 = (-12)^2 = 144$.
\n" ); document.write( "We use the identity
\n" ); document.write( "$$ A^3 + B^3 + C^3 - 3ABC = (A+B+C)(A^2+B^2+C^2 - AB - BC - CA) $$
\n" ); document.write( "$$ A^3 + B^3 + C^3 = (A+B+C)((A+B+C)^2 - 3(AB+BC+CA)) + 3ABC $$
\n" ); document.write( "We have $A+B+C = rs+st+tr = -7$.
\n" ); document.write( "$AB+BC+CA = (rs)(st) + (st)(tr) + (tr)(rs) = rst(r+s+t) = (-12)(4) = -48$.
\n" ); document.write( "$ABC = (rst)^2 = (-12)^2 = 144$.
\n" ); document.write( "Then
\n" ); document.write( "\begin{align*} A^3 + B^3 + C^3 &= (-7)((-7)^2 - 3(-48)) + 3(144) \\ &= -7(49 + 144) + 432 \\ &= -7(193) + 432 \\ &= -1351 + 432 \\ &= -919 \end{align*}
\n" ); document.write( "Therefore,
\n" ); document.write( "$$ S = \frac{A^3+B^3+C^3}{144} = \frac{-919}{144} $$\r
\n" ); document.write( "\n" ); document.write( "Final Answer: The final answer is $\boxed{-919/144}$
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