SOLUTION: Let r, s, and t be solutions of the equation {{{x^3 + 2x^2 - 5x + 15 = 0}}}. Compute {{{1/(r - 2s - 2t) + 1/(s - 2r - 2t) + 1/(t - 2r - 2s)}}}

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Question 1209731: Let r, s, and t be solutions of the equation .
Compute
Found 3 solutions by CPhill, ikleyn, Edwin McCravy:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let the given cubic equation be $x^3 + 2x^2 - 5x + 15 = 0$.
Let $r, s, t$ be the roots of the equation.
By Vieta's formulas, we have:
\begin{align*} \label{eq:1} r + s + t &= -2 \\ rs + st + tr &= -5 \\ rst &= -15\end{align*}
We want to compute
$$ S = \frac{1}{r - 2s - 2t} + \frac{1}{s - 2r - 2t} + \frac{1}{t - 2r - 2s} $$
We can rewrite the denominators as follows:
\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*}
Then,
$$ S = \frac{1}{3r+4} + \frac{1}{3s+4} + \frac{1}{3t+4} $$
Let $y = 3x+4$, so $x = \frac{y-4}{3}$. Substituting this into the cubic equation, we get
$$ \left(\frac{y-4}{3}\right)^3 + 2\left(\frac{y-4}{3}\right)^2 - 5\left(\frac{y-4}{3}\right) + 15 = 0 $$
$$ (y-4)^3 + 6(y-4)^2 - 45(y-4) + 405 = 0 $$
$$ y^3 - 12y^2 + 48y - 64 + 6(y^2 - 8y + 16) - 45y + 180 + 405 = 0 $$
$$ y^3 - 12y^2 + 48y - 64 + 6y^2 - 48y + 96 - 45y + 180 + 405 = 0 $$
$$ y^3 - 6y^2 - 45y + 622 = 0 $$
The roots of this equation are $3r+4$, $3s+4$, and $3t+4$. Let $u = 3r+4$, $v = 3s+4$, $w = 3t+4$.
Then
$$ \frac{1}{u} + \frac{1}{v} + \frac{1}{w} = \frac{uv+vw+uw}{uvw} $$
From the equation $y^3 - 6y^2 - 45y + 622 = 0$, we have
\begin{align*} u+v+w &= 6 \\ uv+vw+uw &= -45 \\ uvw &= -622 \end{align*}
Therefore,
$$ S = \frac{1}{3r+4} + \frac{1}{3s+4} + \frac{1}{3t+4} = \frac{uv+vw+uw}{uvw} = \frac{-45}{-622} = \frac{45}{622} $$
Final Answer: The final answer is $\boxed{45/622}$
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
.

The answer in the post by @CPhill,   ,   is incorrect.

He made errors in his calculations.

The correct answer is   .



Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!

Couldn't find an online solver to help me much on this one.









Write s for r, r for s, and t for t gives 



Write t for r, r for s, and s for t gives 



So we want:





Multiply the numerator out:


Write r for s and t for t:

Write r for s and s for t


Add those three equations term by term:






That's the numerator.

Multiply the denominator out:







That's the denominator.

So the final answer is 

Edwin
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