Thus, the two roots of the equation are \(\omega,\ \omega^2\)

These are well-known by the name of cube roots of unity because when you solve the equation \(x^3 - 1=0\), you get the roots as \(1,\ \omega,\ \omega^2\). And thus, by Vieta's formula, we have

\[1+\omega+\omega^2=0\]

\[\text{&} \qquad \omega^3 = 1\]

Using the above properties, we can easily find that -

good method kishlaya.thanks.but i know this problem has a very simpler way.unfortunately i couldnt remember it .could anyone solve this problem in another way???

Easy Math Editor

`*italics*`

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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## Comments

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TopNewestSolving the first equation, we get

\[x = \frac{-1 \pm \sqrt{3}i}{2}\]

Let us denote \[\omega = \frac{-1 + \sqrt{3}i}{2}\]

Then it is easy to see that

\[\omega^2 = \left(\frac{-1 + \sqrt{3}i}{2}\right)^2 = \frac{-1 - \sqrt{3}i}{2}\]

Thus, the two roots of the equation are \(\omega,\ \omega^2\)

These are well-known by the name of cube roots of unity because when you solve the equation \(x^3 - 1=0\), you get the roots as \(1,\ \omega,\ \omega^2\). And thus, by Vieta's formula, we have

\[1+\omega+\omega^2=0\]

\[\text{&} \qquad \omega^3 = 1\]

Using the above properties, we can easily find that -

\[5x^{234}-x^{99} = 5\omega^{234} - \omega^{99} = 5(\omega^3)^{78} - (\omega^3)^{33} = 5-1 = \boxed{4}\]-

Also, it doesn't matters whichever value of \(x\) you substitute, you'll always get the same answer as you can see -

\[5x^{234}-x^{99} = 5(\omega^2)^{234} - (\omega^2)^{99} = 5(\omega^3)^{156} - (\omega^3)^{66} = 5-1 = \boxed{4}\]

I hope you got my method.

Thanks.

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good method kishlaya.thanks.but i know this problem has a very simpler way.unfortunately i couldnt remember it .could anyone solve this problem in another way???

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Roots of the given equation are \(\Large{w~\&~w^2}\). Hence the answer is 4.

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could you please explain more about it

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