If we square the equation, \(x + \dfrac1x = \sqrt3 \), we get

\[\begin{eqnarray} \left( x + \dfrac1x \right)^2 &=& (\sqrt3)^2 \\ x^2 + \dfrac1{x^2} + 2 &=& 3 \\ x^2 + \dfrac1{x^2} &=& 1 \end{eqnarray} \]

What is the value of the \(x^8 + \dfrac1{x^8} \)?

**Hint**: Try squaring the equation again, what happens?

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