What is $ (3 + w + 2w^2)^4 $ ? where $w$ is cube roots of unity

Question

My progress:
We know that 1 + w + w² = 0, so

(3 -1 - w² + 2w²)⁴
or, (2+ w²)⁴

How can I progress after this. Every suggestion will be appreciated.

Answer 1

Welcome to Mathematics Stack Exchange.

To calculate $(2+w^2)^4$, I would suggested squaring $2+w^2$, simplifying, and squaring the result.

(To simplify, remember that $w^4=w$ and $w^2+w+1=0$.)

Answer 2

Take instead

$$(3+w+2(-1-w))^4=(1-w)^4=1-4w+6w^2-4w^3+w^4=$$

$$=1-4w+6w^2-4+w=1-4w+6(-1-w)-4+w= $$

$$= -9-9w$$

up to you...

Answer 3

We start from $(3+w+2w^2)^4$

As you already mentioned, we know $1+w+w^2 = 0$, so we reduce the starting expression to $(2+w^2)^4$. WIth further substitution, since $w^2=-1-w$, we get:

$(2+w^2)^4 = (2-1-w)^4=(1-w)^4 = 1-4w+6w^2-4w^3+w^4$. Can you continue simplyfying from there?