If α = -1 + -3 2 , then what is the value of (1 + α^ 19 - α^ 35 )^...

Given $\alpha = \dfrac{-1 + \sqrt{-3}}{2}$, which is the complex cube root of unity, $\omega$.

We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

Substituting $\alpha = \omega$ into the given expression, we simplify the powers of $\omega$:

$\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = \omega$

$\alpha^{35} = \omega^{35} = (\omega^3)^{11} \cdot \omega^2 = \omega^2$

$\alpha^{25} = \omega^{25} = (\omega^3)^8 \cdot \omega = \omega$

$\alpha^{38} = \omega^{38} = (\omega^3)^{12} \cdot \omega^2 = \omega^2$

The first term becomes:

$(1 + \alpha^{19} - \alpha^{35})^{100} = (1 + \omega - \omega^2)^{100}$

Since $1 + \omega = -\omega^2$, this reduces to:

$(-\omega^2 - \omega^2)^{100} = (-2\omega^2)^{100} = 2^{100} \omega^{200}$

Since $200 = 3 \times 66 + 2$, $\omega^{200} = \omega^2$. Thus, the first term is $2^{100} \omega^2$.

The second term becomes:

$(1 - 3\alpha^{25} + \alpha^{38})^{50} = (1 - 3\omega + \omega^2)^{50}$

Since $1 + \omega^2 = -\omega$, this reduces to:

$(-\omega - 3\omega)^{50} = (-4\omega)^{50} = 4^{50} \omega^{50} = (2^2)^{50} \omega^{50} = 2^{100} \omega^{50}$

Since $50 = 3 \times 16 + 2$, $\omega^{50} = \omega^2$. Thus, the second term is $2^{100} \omega^2$.

Subtracting the second term from the first term:

$2^{100} \omega^2 - 2^{100} \omega^2 = 0$

Answer: $0$