MHB Evaluate the sum of the reciprocals

[anemone]

Given

$p+q+r+s=0$

$pqrs=1$

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

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

 

[mente oscura]

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.

Spoiler

\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.

 

[MarkFL]

Spoiler

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:

$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}=661$$

 

[I like Serena]

Given

$p+q+r+s=0$

$pqrs=1$

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

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

Spoiler

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 $pqrs$:
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s} = Σpqr = 661 \qquad \blacksquare$$

 

[juantheron]

Spoiler

$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \frac{pqrs}{p}+\frac{pqrs}{q}+\frac{rspq}{r}+\frac{pqrs}{s}$ (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 $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \left(pqr+qrs+rsp+spq\right) = \frac{1983}{3} = 661$

 

[anemone]

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.