If $x,y,z,p,q,r$ are real numbers such that \[\frac{1}{x+p}+\frac{1}{y+p}+\frac{1}{z+p}=\frac{1}{p}\]\[\frac{1}{x+q}+\frac{1}{y+q}+\frac{1}{z+q}=\frac{1}{q}\]\[\frac{1}{x+r}+\frac{1}{y+r}+\frac{1}{z+r}=\frac{1}{r}.\]Find the numerical value of $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$.
Problem
Source: NMTC 2017 Junior P2
Tags: algebra, polynomial
WizardMath
28.10.2017 15:29
$p, q, r$ are the roots of the same cubic equation, so the requested value $\frac{qr+rp+pq}{pqr}$ can be calculated from Vieta's relations on that cubic.
Crazymathsolympiad
29.10.2017 11:48
Is the answer 0
RagvaloD
29.10.2017 11:57
After full expanding we get, that $p,q,r$ are roots of $2X^3+(x^2+y^2+z^2)X^2-xyz=0$ So $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{pq+pr+qr}{pqr}=0$
cowboy_37
20.12.2019 17:59
If x=y=z and p=q=r then 3 is also an answer of this sum
soumyajit_math
02.05.2020 07:58
The answer is 0 First consider p, q and r as the root of a cubic equation.then considering x, y, z as the real numbers, simplify the equation and then the desired answer comes from Vieta's theorem.