Vietnamese Mathematical Olympiad 2022 Let \(a\), \(b\), and \(c\) be the roots of the equation $$ x^3-x^2-1=0. $$ Find the value of $$ \frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}. $$

Let \(a\), \(b\), and \(c\) be nonzero real numbers satisfying

$$ a+b+c=2022, $$ and $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2022}. $$

Find the value of

$$ \frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}. $$

Solution

Solution

Since

$$ a+b+c=2022,\qquad \frac1a+\frac1b+\frac1c=\frac1{2022}, $$

we have

$$ \frac1{a+b+c}=\frac1a+\frac1b+\frac1c. $$

Hence,

$$ \begin{aligned} \frac1{a+b+c}-\frac1a &=\frac1b+\frac1c,\\[2mm] \frac{a-(a+b+c)}{a(a+b+c)} &=\frac{b+c}{bc},\\[2mm] -\frac{b+c}{a(a+b+c)} &=\frac{b+c}{bc}. \end{aligned} $$

Therefore,

$$ \begin{aligned} \frac{b+c}{a(a+b+c)}+\frac{b+c}{bc}&=0,\\[2mm] (b+c)\left(\frac1{a(a+b+c)}+\frac1{bc}\right)&=0,\\[2mm] (b+c)\cdot \frac{a(a+b+c)+bc}{abc(a+b+c)}&=0. \end{aligned} $$

Thus,

$$ (b+c)\bigl(a^2+ab+ac+bc\bigr)=0. $$

Observe that

$$ a^2+ab+ac+bc=(a+b)(a+c). $$

Hence,

$$ (a+b)(b+c)(a+c)=0. $$

There are three possible cases.

Case 1: \(a+b=0\)

Then \(a=-b\). Since \(2023\) is odd,

$$ \frac1{a^{2023}}=-\frac1{b^{2023}}. $$

Also,

$$ a+b+c=2022 \quad\Longrightarrow\quad c=2022. $$

Therefore,

$$ \frac1{a^{2023}} +\frac1{b^{2023}} +\frac1{c^{2023}} = \frac1{2022^{2023}}. $$

Case 2: \(b+c=0\)

Similarly,

$$ a=2022, $$

and hence

$$ \frac1{a^{2023}} +\frac1{b^{2023}} +\frac1{c^{2023}} = \frac1{2022^{2023}}. $$

Case 3: \(a+c=0\)

Likewise,

$$ b=2022, $$

which again gives

$$ \frac1{a^{2023}} +\frac1{b^{2023}} +\frac1{c^{2023}} = \frac1{2022^{2023}}. $$

Answer

$$ \boxed{\frac1{2022^{2023}}.} $$

 

Solution by: Thin Sokkean

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