16. $$ \frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}= $$

Question

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Let $$ \omega $$ be a nonreal cube root of unity so that

$$
\omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0.
$$

We begin with the expression

$$
\frac{a + b \omega + c \omega^2}{c + a \omega + b \omega^2}.
$$

Notice that if we multiply the numerator $$ (a + b \omega + c \omega^2) $$ by $$ \omega $$, we get

$$
\omega (a + b \omega + c \omega^2) = a \omega + b \omega^2 + c \omega^3.
$$

Since $$ \omega^3 = 1 $$, this simplifies to

$$
a \omega + b \omega^2 + c.
$$

Observing the denominator, we see that

$$
c + a \omega + b \omega^2 = a \omega + b \omega^2 + c.
$$

Thus, we have established that

$$
\omega (a + b \omega + c \omega^2) = c + a \omega + b \omega^2.
$$

This allows us to rewrite the original expression as

$$
\frac{a + b \omega + c \omega^2}{\omega (a + b \omega + c \omega^2)} = \frac{1}{\omega}.
$$

Since for a cube root of unity $$ \frac{1}{\omega} = \omega^2 $$, the expression simplifies to

$$
\omega^2.
$$
The expression simplifies to $$ \omega^2 $$.

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