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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 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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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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