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If \(x+\dfrac{1}{x}=1\) ,then the value of \(x^{12}+x^9+x^6+x^3+1\) is?

Note by Gautam Arya 8 months, 3 weeks ago

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\(\begin{align} x+\dfrac{1}{x}&=1 \hspace{5mm}\color{blue}\text{given}\\ \implies x^2-x+1&=0 \\\\ \text{Multiplying both sides} &\text{with x+1,we get,}\\ x^3+1&=0\\ \implies x^3&=-1\\ \text{Substituting for}&x^3 \text{we get}\\ 1+x^3+x^6+x^9+x^{12}&=1-1+1-1+1=\boxed{1}\end{align} \)

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

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

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`[example link](https://brilliant.org)`

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewest\(\begin{align} x+\dfrac{1}{x}&=1 \hspace{5mm}\color{blue}\text{given}\\ \implies x^2-x+1&=0 \\\\ \text{Multiplying both sides} &\text{with x+1,we get,}\\ x^3+1&=0\\ \implies x^3&=-1\\ \text{Substituting for}&x^3 \text{we get}\\ 1+x^3+x^6+x^9+x^{12}&=1-1+1-1+1=\boxed{1}\end{align} \)

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