If `a+b+c=2022` and `1/a+1/b+1/c=1/2022` Find the value of `1/(a^2023)+1/(b^2023)+1/(c^2023)`
Solution
We had: `a+b+c=2022` and `1/a+1/b+1/c=1/2022`
Then, `1/(a+b+c)=1/2022` So that, `1/(a+b+c)=1/a+1/b+1/c`
`\rightarrow1/(a+b+c)-1/a=1/b+1/c`
`\leftrightarrow(a-a-b-c)/(a(a+b+c))=(b+c)/(bc)`
`leftrightarrow-(b+c)/(a(a+b+c))=(b+c)/(bc)`
`leftrightarrow(b+c)/(a(a+b+c))+(b+c)/(bc)`
`leftrightarrow(b+c)(1/(a(a+b+c))+1/(bc))=0`
`leftrightarrow(b+c)((a(a+b+c)+bc)/(abc(a+b+c)))=0`
Therefore, `(b+c)(bc+a^2+ab+ac)=0`
`leftrightarrow(b+c)[c(b+a)+a(a+b)]=0`
`leftrightarrow(a+b)(b+c)(a+c)=0`
There are 3 cases of above equatoin:
+ Case 01: `a+b=0` Then `a=-b` It is easy to see that `1/a^2023=-1/b^2023`
Also from this `a+b+c=2022` We can see that `c=2022`
Then, `1/c^2023=1/2022^2023`
Hence, `1/(a^2023)+1/(b^2023)+1/(c^2023)=1/2022^2023`
+ For other 2 cases `b+c=0` and `a+c=0` We do the same and we will get the answer is `1/(a^2023)+1/(b^2023)+1/(c^2023)=1/2022^2023`
Solution by: Thin Sokkean
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