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The stated duplicate question does not address solution in terms of $p, q, r,$ i.e. not in terms of the Viete's formula. My approach is based on the viete's formula by finding the relation between the coefficients of a polynomial to sums and products of its roots.

If $\frac 1{a+b+c}= \frac 1a+ \frac 1b+ \frac 1c$, then show for odd $n$, $ \frac 1{a^n+b^n+c^n}= \frac 1{a^n}+ \frac 1{b^n}+ \frac 1{c^n}$

Cannot establish the significance of odd $n$, i.e., why not applicable for even $n$.

Also, a simple approach is: $\frac 1{a+b+c}= \frac{ab+bc+ca}{abc}\implies abc = (ab+bc+ca)(a+b+c)$.

The two product terms are usually associated to a polynomial (as shown below) with $3$ roots: $a,b,c,$ having $-p=$ sum of roots $=a+b+c$, and $q=$ sum of product of roots two at a time$=ab+bc+ca$, and the single term for all $3$ roots $=-r= abc.$

So, $-r = -pq\implies r = pq$.

Derivation of polynomial with $3$ roots:
Let $a,b,c$ be the roots of $x^3 + px^2 + qx + r = 0$. Then, writing the polynomial in terms of roots gives $(x - a)(x - b)(x -c)$.

$\therefore$ $x^3 + px^2 + qx + r = (x - a)(x - b)(x - c)$.

$= (x^2 - (a + b)x + ab)(x - c)$

$= x^3 - (a + b)\cdot x^2 + (ab)\cdot x - c\cdot x^2 + (a+b)c\cdot x - abc$

$= x^3 - (a + b +c) x^2 + (ab + bc + ca) x - abc$

$\therefore$ equating coefficients

(a) $a + b + c = -p$.

(b) $ab+ bc + ca = q$.

(c) $abc = -r$.

Unable to proceed further.

I want to add that the source has cryptic answer that has next step as :
The required polynomial is : $f(x)= x^3 +px^2+qx+pq$. <-- This step stucks me.
The next one ones are easy:
This implies $(x+p)(x^2+q)$.
As one root is $=-p$, so $a=-p= a+b+c\implies c=-b$.
So, $\frac 1{a^n+b^n+c^n} = \frac 1{a^n}=\frac 1{a^n+b^n-b^n}=\frac 1{a^n+b^n+c^n}$.

The last part uses odd $n$ property, but the stucking step is issue.

jiten
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  • Are $a,b,c$ natural or real (with appropriate assumptions so you don't divide by zero)? – Cameron Williams Jun 02 '18 at 03:50
  • @CameronWilliams It is not given for $a,b,c$ only the odd value of $n$ is stated that implies $n$ being an integer. – jiten Jun 02 '18 at 03:58
  • @AlexFrancisco Wrong, as the solution approach in the single answer there is not using Viete's formula, as my approach is. Approach should matter too. – jiten Jun 02 '18 at 04:04
  • @jiten If your method doesn't work, why stick to it? – Ѕᴀᴀᴅ Jun 02 '18 at 04:12
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    But, have seen solution somewhere, but unable to understand it. I took the Viete's approach of the cryptic solution, but failed in reaching on my own the last part (as shown). For your reference, the next steps given in the stated (cryptic) solution are : The required polynomial is : $f(x)= x^3 +px^2+qx+pq$. <<This step stucks me. The next one ones are easy: This implies $(x+p)(x^2+q)$. As one root is $=-p$, so $a=-p= a+b+c\implies c=-b$. So, $\frac 1{a^n+b^n+c^n} = \frac 1{a^n}=\frac 1{a^n+b^n-b^n}=\frac 1{a^n+b^n+c^n}$. The last part uses odd $n$ property, but the stucking step is issue. – jiten Jun 02 '18 at 04:24
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    This question is not a duplicate. The OP wants to know how to use Vieta's Formulas to do it. – Cheerful Parsnip Jun 02 '18 at 04:51
  • @jiten: Now that you know that your question is very similar to an existing question (so similar in fact that $4$ out of the required $5$ users voted to close it as a duplicate), I would suggest to edit your question to a) link to the similar question so that insights can be shared between the two threads and b) state more clearly that your question is about solving the problem using Vieta's formulas to solve the problem and not about solving it in general. (That isn't clear to me from reading it, even after you mentioned this aspect in the first line.) – joriki Jun 02 '18 at 05:05
  • @joriki Hope that the edit suffices. Specified link, as well as stressed my approach's difference. – jiten Jun 02 '18 at 05:26
  • It's better, but what's confusing about it (and probably caused $4$ people to vote to close) is that you talk about Vieta's formula being your approach. If this is merely your approach to solving the problem, then the other question solves your problem, just with a different approach. The reason for keeping open this question would be if your problem is to solve this using Vieta's formula. – joriki Jun 02 '18 at 08:08

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The polynomial is $$f(x)=x^3+px^2+qx+r=(x-a)(x-b)(x-c)$$

You have proven that $r=pq$ from the condition that is given.

$$(x+p)(x^2+q)=(x-a)(x-b)(x-c)$$

$$(x+p)(x-i\sqrt{q)}(x+i\sqrt{q})=(x-a)(x-b)(x-c)$$

WLOG, if $a=-p$, then $b=-c$ and $\frac1{b}=-\frac1{c}$.

If $n$ is odd, then we have $b^n=(-1)^nc^n=-c^n$ and $\frac1{b^n}=(-1)^n\frac1{c^n}=-\frac1{c^n}$, that is we have $b^n+c^n=0$ and $\frac1{b^n}+\frac1{c^n}=0$. Both sides of the equation that we want to show become $\frac1{a^n}$.

However, if $n$ is even, then we have $b^n=c^n$ and $\frac1{b^n}=\frac1{c^n}$ and the LHS of what we want to show become $\frac1{a^n+2b^n}$ but the RHS becomes $\frac1{a^n}+\frac2{b^n}$.

Siong Thye Goh
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  • Sorry for not able to use the derived fact $r =pq$ in the equation : $x^3 +px^2 +qx+r$. One more thing, from which situation this equation could have been derived, say any geometrical one, or in sequences, etc. I am asking to further link such questions to their origin i.e. where they could have arisen. – jiten Jun 02 '18 at 09:01
  • Please see my post at : https://math.stackexchange.com/q/2805276/424260. – jiten Jun 02 '18 at 10:48