MHB Evaluate the sum of the reciprocals

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The discussion revolves around evaluating the sum of the reciprocals of four variables, given specific conditions: their sum is zero, their product is one, and the sum of their cubes is 1983. Participants express appreciation for the collaborative problem-solving approach, highlighting similarities in their methods. The focus remains on deriving the value of the expression based on the provided equations. The challenge emphasizes the importance of understanding polynomial relationships and symmetric sums. Ultimately, the goal is to find a solution that satisfies all given conditions.
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Given

$p+q+r+s=0$

$pqrs=1$

$p^3+q^3+r^3+s^3=1983$

Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.
 
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anemone said:
Given

$p+q+r+s=0$

$pqrs=1$

$p^3+q^3+r^3+s^3=1983$

Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.

Hello.

\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=

=qrs+prs+pqs+pqr=

=qrs+prs+rrs+srs-rrs-srs+pqs+pqr=

=-rrs-srs+pqs+pqr, (*)

(p+q)^3=-(r+s)^3

p^3+3p^2q+3pq^2+q^3=-r^3-3r^2s-3rs^2-s^3

1983+3p^2q+3pq^2=-3r^2s-3rs^2

661+p^2q+pq^2=-r^2s-rs^2, (**)

For (*) and (**):

\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=

=661+p^2q+pq^2+pqs+pqr=

=661+pq(p+q+s+r)=661

Regards.
 
I would first combine terms in the expression we are asked to evaluate:

$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}= \frac{qrs+prs+pqs+pqr}{pqrs}$$

Since $$pqrs=1$$, we may write:

$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}=qrs+prs+pqs+pqr$$

Next, take the first given equation and cube it to obtain:

$$(p+q+r+s)^3=0$$

This may be expanded and arranged as:

$$-2\left(p^3+q^3+r^3+s^3 \right)+ 6(qrs+prs+pqs+pqr)+ 3(p+q+r+s)\left(p^2+q^2+r^2+s^2 \right)=0$$

Since $p+q+r+s=0$ and $p^3+q^3+r^3+s^3=1983$, we obtain:

$$-2\cdot1983+6\left(qrs+prs+pqs+pqr \right)=0$$

$$qrs+prs+pqs+pqr=\frac{1983}{3}=661$$

And so we may therefore conclude:

$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}=661$$
 
anemone said:
Given

$p+q+r+s=0$

$pqrs=1$

$p^3+q^3+r^3+s^3=1983$

Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.

Now that I have just explored Newton's Identities, this is fun. ;)

Let's define $Σ$ such that $Σp^3 = p^3+q^3+r^3+s^3$.
And for instance $Σpqr = pqr + pqs + prs + qrs$.

Then from Newton's Identies we have:
$$Σp^3 = ΣpΣp^2 - ΣpqΣp + 3Σpqr$$
Since $Σp = 0$, this simplifies to:
$$Σp^3 = 3Σpqr = 1983$$
Therefore:
$$Σpqr = 661$$

Since $pqrs=1$, we get by multiplying with $pqrs$:
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s} = Σpqr = 661 \qquad \blacksquare$$
 
$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \frac{pqrs}{p}+\frac{pqrs}{q}+\frac{rspq}{r}+\frac{pqrs}{s}$ (using $pqrs = 1$)

So $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \left(pqr+qrs+rsp+spq\right)$

Given $p+q+r+s = 0\Rightarrow (p+q)^3 = -(r+s)^3\Rightarrow p^3+q^3+3pq(p+q) = r^3+s^3+3rs(r+s)$

again using $p+q=-(r+s)$ and $(r+s) = -(p+q)$

So we get $p^3+q^3+r^3+s^3 = 3\left(pqr+qrs+rsp+spq\right)$

Given $1983 = 3\left(pqr+qrs+rsp+spq\right)$

So $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \left(pqr+qrs+rsp+spq\right) = \frac{1983}{3} = 661$
 
Thanks to mente oscura, MarkFL, I like Serena and jacks for participating and it feels so great to receive so many replies to my challenge problem and my way of attacking it is exactly the same as jacks's solution.:o
 
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