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Question Number 196714 by mr W last updated on 30/Aug/23

Answered by som(math1967) last updated on 30/Aug/23

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$$$\Rightarrow (a+b+c)(ab+bc+ca)-abc=0$$$$\Rightarrow (a+b)(b+c)(c+a)=0$$$$a+b=0\Rightarrow a=-b\Rightarrow c=2022$$$$\frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}$$$$\Rightarrow \frac{1}{a^{2023}}-\frac{1}{a^{2023}}+\frac{1}{{\left(2022\right)}^{2023}}$$$$\Rightarrow \frac{1}{{\left(2022\right)}^{2023}}$$

Commented by mr W last updated on 30/Aug/23

$$verynice!$$

Answered by Rasheed.Sindhi last updated on 30/Aug/23

$$a+b+c=2022,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2022}$$$$\frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}=?$$$$\frac{a+b}{ab}=\frac{1}{2022}-\frac{1}{c}$$$$=\frac{1}{2022}-\frac{1}{2022-(a+b)}$$$$=\frac{2022-(a+b)-2022}{2022(2022-(a+b))}$$$$=\frac{a+b}{2022((a+b)-2022)}$$$$\frac{a+b}{ab}-\frac{a+b}{2022((a+b)-2022)}=0$$$$(a+b)(\frac{1}{ab}-\frac{1}{2022((a+b)-2022)})=0$$$$a=-b{}^{★}\mid \frac{1}{2022((a+b)-20222)}=\frac{1}{ab}$$$$2022(a+b)-{2022}^2-ab=0$$$$(2022-a)(b-2022)=0$$$$a=2022\mid b=2022$$$$a+b+c=2022$$$$2022+2022+c=2022$$$$c=-2022$$$$\frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}$$$$=\frac{1}{{2022}^{2023}}+\frac{1}{{2022}^{2023}}+\frac{1}{{(-2022)}^{2023}}$$$$=\frac{1}{{2022}^{2023}}+\frac{1}{{2022}^{2023}}-\frac{1}{{2022}^{2023}}$$$$=\frac{1}{{2022}^{2023}}✓$$$${}^{★}a=-b$$$$a+b+c=2022\Rightarrow -b+b+c=2022$$$$\Rightarrow c=2022$$$$\frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}$$$$\frac{1}{{(-b)}^{2023}}+\frac{1}{b^{2023}}+\frac{1}{{\left(2022\right)}^{2023}}$$$$=\frac{1}{{\left(2022\right)}^{2023}}✓$$$$$$

Commented by mr W last updated on 30/Aug/23

$$thanks!$$

Answered by Rasheed.Sindhi last updated on 30/Aug/23

$$a+b+c=2022,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2022}$$$$\frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}=?$$$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2022}$$$$\Rightarrow ab+bc+ca=\frac{abc}{2022}$$$$Leta,b,carerootsofanequation.$$$$Equationwillbe:$$$$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$$$$x^3-2022x^2+\left(\frac{abc}{2022}\right)x-abc=0$$$$2022x^3-{2022}^2{x}^2+abcx-2022abc=0$$$$2022x^2(x-2022)+abc(x-2022)=0$$$$(x-2022)(2022x^2+abc)=0$$$$x=2022\mid 2022x^2+abc=0$$$$leta=2022\mid 2022x^2+2022bc=0$$$$\mid x^2=-bc$$$$\mid x=\pm \sqrt{-bc}$$$$\mid Letb=\sqrt{-bc}\mathrm{\&}c=-\sqrt{-bc}$$$$b+c=0\Rightarrow c=-b$$$$\frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}$$$$=\frac{1}{{2022}^{2023}}+\frac{1}{{\left(b\right)}^{2023}}+\frac{1}{{(-b)}^{2023}}$$$$=\frac{1}{{2022}^{2023}}+\cancel{\frac{1}{{\left(b\right)}^{2023}}}-\cancel{\frac{1}{{\left(b\right)}^{2023}}}$$$$=\frac{1}{{2022}^{2023}}✓$$

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