Given \(x + \dfrac{1}{x} = 4\), find the value of \(x^6 + \dfrac{1}{x^6} - 3\!\left(x^4 + \dfrac{1}{x^4}\right) + 4\!\left(x^2 + \dfrac{1}{x^2}\right)\).
2176
Let \(s = x + \frac{1}{x} = 4\).
Step 1: \(x^2 + \frac{1}{x^2} = s^2 - 2 = 16 - 2 = 14\)
Step 2: \(x^4 + \frac{1}{x^4} = \left(x^2+\frac{1}{x^2}\right)^2 - 2 = 196 - 2 = 194\)
Step 3: \(x^3 + \frac{1}{x^3} = s^3 - 3s = 64 - 12 = 52\)
Step 4: \(x^6 + \frac{1}{x^6} = \left(x^3+\frac{1}{x^3}\right)^2 - 2 = 2704 - 2 = 2702\)
Step 5 – Substitute:
\(2702 - 3(194) + 4(14) = 2702 - 582 + 56 = \mathbf{2176}\)
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