For some \( a, b \), let \( f(x) = \left| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} \right| \), where \( x \neq 0 \), \( \lim_{x \to 0} f(x) = \lambda + \mu a + \nu b \).
Then \( (\lambda + \mu + \nu)^2 \) is equal to:
First, compute the determinant of the matrix as \( x \to 0 \) and then take the limit to find the value of \( \lambda + \mu + \nu \). The limit and determinant calculation gives the value 3 for \( \lambda + \mu + \nu \), so squaring this gives 9.
Final Answer: \( (\lambda + \mu + \nu)^2 = 9 \).
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.