Question Number 167137 by mr W last updated on 07/Mar/22
Answered by som(math1967) last updated on 07/Mar/22
![(1/a)+(1/b) +(1/c)=(1/(a+b+c)) [∵ a+b+c=2022] (a+b+c)(ab+bc+ca)−abc=0 (a+b)(b+c)(c+a)=0 if a+b=0 ∴ c=2022 (1/a^(2023) )+(1/b^(2023) )+(1/c^(2023) )=0+((1/(2022)))^(2023) [for b+c=0 or c+a=0 gives same result]](https://www.tinkutara.com/question/Q167138.png)
$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\:+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}+{b}+{c}}\:\:\left[\because\:{a}+{b}+{c}=\mathrm{2022}\right] \\ $$$$\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−{abc}=\mathrm{0} \\ $$$$\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)=\mathrm{0} \\ $$$${if}\:{a}+{b}=\mathrm{0} \\ $$$$\therefore\:{c}=\mathrm{2022} \\ $$$$\:\frac{\mathrm{1}}{{a}^{\mathrm{2023}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2023}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2023}} }=\mathrm{0}+\left(\frac{\mathrm{1}}{\mathrm{2022}}\right)^{\mathrm{2023}} \\ $$$$\left[{for}\:{b}+{c}=\mathrm{0}\:{or}\:{c}+{a}=\mathrm{0}\:{gives}\right. \\ $$$$\left.\:{same}\:{result}\right] \\ $$
Commented by mr W last updated on 07/Mar/22
$${thanks}\:{sir}! \\ $$