SOLUTION: Let r, s, and t be solutions of the equation x^3 - 4x^2 - 7x + 12 = 0. Compute \frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2}

Question 1209730: Let r, s, and t be solutions of the equation x^3 - 4x^2 - 7x + 12 = 0. Compute
\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2}
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $r, s, t$ be the roots of the cubic equation $x^3 - 4x^2 - 7x + 12 = 0$.
By Vieta's formulas, we have:
\begin{align*} r+s+t &= 4 \\ rs+st+tr &= -7 \\ rst &= -12 \end{align*}
We want to compute
$$ S = \frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2} = \frac{r^3s^3 + r^3t^3 + s^3t^3}{r^2s^2t^2} $$
We know that $rst = -12$, so $r^2s^2t^2 = (rst)^2 = (-12)^2 = 144$.
Let $A = rs$, $B = st$, and $C = tr$. Then we want to compute
$$ \frac{A^3 + B^3 + C^3}{144} $$
We have $A+B+C = rs+st+tr = -7$.
Also, $ABC = (rs)(st)(tr) = (rst)^2 = (-12)^2 = 144$.
We use the identity
$$ A^3 + B^3 + C^3 - 3ABC = (A+B+C)(A^2+B^2+C^2 - AB - BC - CA) $$
$$ A^3 + B^3 + C^3 = (A+B+C)((A+B+C)^2 - 3(AB+BC+CA)) + 3ABC $$
We have $A+B+C = rs+st+tr = -7$.
$AB+BC+CA = (rs)(st) + (st)(tr) + (tr)(rs) = rst(r+s+t) = (-12)(4) = -48$.
$ABC = (rst)^2 = (-12)^2 = 144$.
Then
\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*}
Therefore,
$$ S = \frac{A^3+B^3+C^3}{144} = \frac{-919}{144} $$
Final Answer: The final answer is $\boxed{-919/144}$

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