# Evaluate the sum of the reciprocals

#### anemone

##### MHB POTW Director
Staff member
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

##### Well-known member
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.

#### MarkFL

Staff member
I would first combine terms in the expression we are asked to evaluate:

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

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

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

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

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

This may be expanded and arranged as:

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

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

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

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

And so we may therefore conclude:

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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 $pqrs$:
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s} = Σpqr = 661 \qquad \blacksquare$$

#### jacks

##### Well-known member
$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \frac{pqrs}{p}+\frac{pqrs}{q}+\frac{rspq}{r}+\frac{pqrs}{s}$ (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

##### MHB POTW Director
Staff member
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.