Question Number 190523 by Rupesh123 last updated on 04/Apr/23
Answered by mr W last updated on 01/Sep/23
![(m+n+p)(mn+np+pm)=mnp (m+n)(mn+np+pm+p^2 )=0 (m+n)[m(n+p)+(n+p)p]=0 (m+n)(n+p)(p+m)=0 ⇒m+n=0 (1/((m+n+p)^(2923) ))−(1/m^(2023) )−(1/n^(2023) )−(1/p^(2923) ) =(1/p^(2923) )+(1/n^(2023) )−(1/n^(2023) )−(1/p^(2923) ) =0](https://www.tinkutara.com/question/Q196854.png)
$$\left({m}+{n}+{p}\right)\left({mn}+{np}+{pm}\right)={mnp} \\ $$$$\left({m}+{n}\right)\left({mn}+{np}+{pm}+{p}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left({m}+{n}\right)\left[{m}\left({n}+{p}\right)+\left({n}+{p}\right){p}\right]=\mathrm{0} \\ $$$$\left({m}+{n}\right)\left({n}+{p}\right)\left({p}+{m}\right)=\mathrm{0} \\ $$$$\Rightarrow{m}+{n}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\left({m}+{n}+{p}\right)^{\mathrm{2923}} }−\frac{\mathrm{1}}{{m}^{\mathrm{2023}} }−\frac{\mathrm{1}}{{n}^{\mathrm{2023}} }−\frac{\mathrm{1}}{{p}^{\mathrm{2923}} } \\ $$$$=\frac{\mathrm{1}}{{p}^{\mathrm{2923}} }+\frac{\mathrm{1}}{{n}^{\mathrm{2023}} }−\frac{\mathrm{1}}{{n}^{\mathrm{2023}} }−\frac{\mathrm{1}}{{p}^{\mathrm{2923}} } \\ $$$$=\mathrm{0} \\ $$