Given that x + \frac{1}{x} = 3, find the value of $x^6 + \frac{1}{x^6} - 2\left(x^4 + \frac{1}{x^4}\right) + 3\left(x^2 + \frac{1}{x^2}\right)$.

Question

Accepted Answer

249. From $x + \frac{1}{x} = 3$, we get $x^2 + \frac{1}{x^2} = 3^2 - 2 = 7$. Then $x^4 + \frac{1}{x^4} = 7^2 - 2 = 47$ and $x^6 + \frac{1}{x^6} = 7\cdot 47 - 7 = 329$ using standard identities. Substituting gives 329 - 2(47) + 3(7) = 249.

Given that x + \frac{1}{x} = 3, find the value of \(x^6 + \frac{1}{x^6} - 2\left(x^4 + \frac{1}{x^4}\right) + 3\left(x^2 + \frac{1}{x^2}\right)\).

Solution

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