Given: \(x + \dfrac{1}{x} = -2\), then determine the value of \(x^6 + \dfrac{1}{x^6} - 3\!\left(x^4 + \dfrac{1}{x^4}\right) + 4\!\left(x^2 + \dfrac{1}{x^2}\right)\)
4
From \(x + \dfrac{1}{x} = -2\) we can either solve for \(x\) directly or build up the powers.
Direct route: Multiply by \(x\): \(x^2 + 2x + 1 = 0 \;\Longrightarrow\; (x+1)^2 = 0 \;\Longrightarrow\; x = -1\).
For \(x = -1\): \(x^2 = 1,\ x^4 = 1,\ x^6 = 1\) and \(\dfrac{1}{x^2}=\dfrac{1}{x^4}=\dfrac{1}{x^6}=1\).
So \(x^2 + 1/x^2 = 2\), \(x^4 + 1/x^4 = 2\), \(x^6 + 1/x^6 = 2\).
Substitute:
\(\text{Expr} = 2 - 3(2) + 4(2) = 2 - 6 + 8 = 4\)
Hence the value is 4 — option (4).
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