Problem Statement
Let $\omega$ be a complex non-real cube root of unity, so that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
(a) Simplify $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$. [3 Marks]
(b) Prove that $(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5 = 32$. [3 Marks]
Verified Solution & Marking Scheme
Substitute 1 + ω² = -ω and 1 + ω = -ω²
From $1 + \omega + \omega^2 = 0$, we have $1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$.
$$(1 - \omega + \omega^2) = (1 + \omega^2) - \omega = -\omega - \omega = -2\omega$$
$$(1 + \omega - \omega^2) = (1 + \omega) - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$$
Part (a) Product Evaluation
$$(1 - \omega + \omega^2)(1 + \omega - \omega^2) = (-2\omega)(-2\omega^2) = 4\omega^3$$
Since $\omega^3 = 1$, the product is $4(1) = 4$.
Part (b) Fifth Powers Evaluation
$$(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5 = (-2\omega)^5 + (-2\omega^2)^5$$
$$= -32\omega^5 - 32\omega^{10}$$
Since $\omega^3 = 1$, we have $\omega^5 = \omega^3 \cdot \omega^2 = \omega^2$ and $\omega^{10} = (\omega^3)^3 \cdot \omega = \omega$.
$$= -32(\omega^2 + \omega)$$
Since $1 + \omega + \omega^2 = 0 \implies \omega^2 + \omega = -1$:
$$= -32(-1) = 32 \quad \text{(Q.E.D.)}$$