If $1, \omega, \omega^2$ are cube roots, show that $(2-\omega)(2-\omega^2)(2-\omega^{19})(2-\omega^{23})=49$

Question

If $1, \omega, \omega^2$ are cube roots, how to show that

$$ (2-\omega) \left(2-\omega^2\right) \left(2-\omega^{19}\right) \left(2-\omega^{23}\right) = 49$$

I really don't know how to start. Help!

Answer 1

Clearly $\omega^3=1$ and $1+\omega+\omega^2=0$. So, the problem reduces to:

$$(2-\omega)(2-\omega^2)(2-\omega)(2-\omega^{2})=(2-\omega)^2(2-\omega^2)^2=(4+\omega^2-4\omega)(4+\omega^4-4\omega^2)$$ $$=(4+\omega^2-4\omega)(4+\omega-4\omega^2)=16+4\omega-16\omega^2+4\omega^2+1-4\omega-16\omega-4\omega^2+16$$ $$=33+-16(\omega+\omega^2)=\boxed{49}$$

since $\omega+\omega^2=-1$.

Answer 2

Using modular arhitmetic and the fact that $\omega^3=1$, we can write: $$(2-\omega) \left(2-\omega^2\right) \left(2-\omega^{19}\right) \left(2-\omega^{23}\right) =(5-2\omega^2-2\omega)\cdot(5-2\omega^2-2\omega)$$ Now, rearranging this expression and using the fact that $1+\omega+\omega^2=0$, we arrive at: $$(2-\omega) \left(2-\omega^2\right) \left(2-\omega^{19}\right) \left(2-\omega^{23}\right) = 49$$

Answer 3

Since $\omega^3=1$ and $\omega^2+\omega+1=0$, we have $$ \begin{align} (2-\omega)(2-\omega^2)(2-\omega^{19})(2-\omega^{23}) &= (2-\omega)(2-\omega^2)(2-\omega^{1})(2-\omega^{2}) \\&= ((2-\omega)(2-\omega^2))^2 \\&= (-2 \omega^2 - 2 \omega + 5)^2 \\&= (-2(\omega^2 + \omega) + 5)^2 \\&= (-7)^2 =49 \end{align} $$

Answer 4

Clearly we have that $1, \omega, \omega^2$ are the roots of the polynomial $p(x):=x^3-1$. Therefore, it follows that $$p(x)=(x-1)(x-\omega)(x-\omega^2)$$ and that $\omega^3=1$. Now, $$\begin{align}(2-\omega)(2-\omega^2)(2-\omega^{19})(2-\omega^{23})&=(2-1)(2-\omega)(2-\omega^2)(2-1)(2-\omega)(2-\omega^2)\\&=p(2)^2\\&=49\end{align}$$

Answer 5

Since $\omega,\omega^{2}$ are the roots of $f(x)=x^{2}+x+1$,

$$ (2-\omega)(2-\omega^{2})(2-\omega^{19})(2-\omega^{23}) =\left(f(2)\right)^{2}=49 $$