
From 1st lens
1/v + 1/6 = 1/24
1/v = 1/24 - 1/6 = -1/8
v = -8 cm
From 2nd lens
1/v + 1/18 = 1/9
1/v = 1/9 - 1/18 = 1/18
v = 18 cm
So distance between object and its image:
6 + 10 + 18 = 34 cm
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 ___________.