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.
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 $\frac1{a^n}+\frac2{b^n}$.