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Evaluate the sum of the reciprocals

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
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
Nov 29, 2013
172
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.

[tex]\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=[/tex]

[tex]=qrs+prs+pqs+pqr=[/tex]

[tex]=qrs+prs+rrs+srs-rrs-srs+pqs+pqr=[/tex]

[tex]=-rrs-srs+pqs+pqr[/tex], (*)

[tex](p+q)^3=-(r+s)^3[/tex]

[tex]p^3+3p^2q+3pq^2+q^3=-r^3-3r^2s-3rs^2-s^3[/tex]

[tex]1983+3p^2q+3pq^2=-3r^2s-3rs^2[/tex]

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

For (*) and (**):

[tex]\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=[/tex]

[tex]=661+p^2q+pq^2+pqs+pqr=[/tex]

[tex]=661+pq(p+q+s+r)=661[/tex]



Regards.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Mar 5, 2012
8,780
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
Apr 5, 2012
226
$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \frac{pqrs}{p}+\frac{pqrs}{q}+\frac{rspq}{r}+\frac{pqrs}{s}$ (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 $\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = \left(pqr+qrs+rsp+spq\right) = \frac{1983}{3} = 661$
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
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.:eek: