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

Question 1209738: Let r, s, and t be solutions of the equation 3x^3 - 4x^2 - 2x + 12 = 0. Compute
\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2}.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
1. **Vieta's Formulas:**
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:
* rst = -d/a = -12/3 = -4
* rs + rt + st = c/a = -2/3
2. **Rewrite the Expression:**
We want to compute:
(rs/t²) + (rt/s²) + (st/r²) = (r³s³ + r³t³ + s³t³) / (r²s²t²)
3. **Simplify using Vieta's Formulas:**
We know that r²s²t² = (rst)² = (-4)² = 16. So, we need to find r³s³ + r³t³ + s³t³.
4. **Key Identity:**
Recall the identity: A³ + B³ + C³ - 3ABC = (A + B + C)(A² + B² + C² - AB - BC - CA)
Let A = rs, B = rt, and C = st. Then:
(rs)³ + (rt)³ + (st)³ - 3(rst)² = (rs + rt + st)[(rs)² + (rt)² + (st)² - rs*rt - rs*st - rt*st]
We have rs + rt + st = -2/3 and rst = -4. Substituting these values:
r³s³ + r³t³ + s³t³ - 3(-4)² = (-2/3)[(rs)² + (rt)² + (st)² - r²st - rs²t - rst²]
r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - rt(rs + st + rt)]
r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - (-4)(-2/3)]
r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - 8/3]
We also know that (rs + rt + st)² = (rs)² + (rt)² + (st)² + 2rst(r + s + t).
(-2/3)² = (rs)² + (rt)² + (st)² + 2(-4)(4/3)
4/9 = (rs)² + (rt)² + (st)² - 32/3
(rs)² + (rt)² + (st)² = 4/9 + 96/9 = 100/9
Now plug this back into the equation:
r³s³ + r³t³ + s³t³ - 48 = (-2/3)[100/9 - 8/3]
r³s³ + r³t³ + s³t³ - 48 = (-2/3)[100/9 - 24/9]
r³s³ + r³t³ + s³t³ - 48 = (-2/3)(76/9)
r³s³ + r³t³ + s³t³ = 48 - 152/27
r³s³ + r³t³ + s³t³ = (1296 - 152)/27 = 1144/27
5. **Final Calculation:**
(r³s³ + r³t³ + s³t³) / (r²s²t²) = (1144/27) / 16 = 1144 / (27 * 16) = 1144 / 432 = 143/54
Final Answer: The final answer is $\boxed{143/54}$

Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.
Let r, s, and t be solutions of the equation 3x^3 - 4x^2 - 2x + 12 = 0.
Compute (rs)/t^2 + (rt)/s^2 + (st)/r^2.
~~~~~~~~~~~~~~~~~~~~

The solution by @CPhill to this problem is INCORRECT.

The error is when he transforms the formula

        r³s³ + r³t³ + s³t³ - 3(-4)² = (-2/3)[(rs)² + (rt)² + (st)² - r²st - rs²t - rst²]

into formula

        r³s³ + r³t³ + s³t³ - 48 = (-2/3)[(rs)² + (rt)² + (st)² - rt(rs + st + rt)]

Indeed, the previous formula is symmetric relative r, s, t, while the last formula is not symmetric.

Hence, everything which follows makes no sense and is irrelevant.

        Honestly, I don't know what is the sense in solving such problems.

The idea is always the same: use the Vieta's formulas.
If you know this idea, the rest is simply arithmetic and many-line transformations of formulas.

Such problems are useful and educative in low degrees.
In high degrees they become non-sensical tedious exercises.

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