Question

if w is the complex cube root of unity then find the value of (1-w-w^2)^3 + (1-w+w^2)^3​

Answer 1

Answer:

0

Step-by-step explanation:

If w is the cube root of unity then

w^3 = 1  and 1 + w + w^2 =0

$(1-w-w^2)^3+(1-w+w^2)^3\\=(1-(w+w^2))^3+((1+w^2)-w)^3\\=(1-(-1))^3+(-w-w)^3\\=2^3+(-2)^3w^3\\=8-8\\=0$

Hope it Helps

Answer 2

$(1-\omega-\omega^{2})^{3}+(1-\omega+\omega^{2})^{3}=0$

Step-by-step explanation:

Since $\omega$ is the complex cube root of unity, then we have two relations:

• $\omega^{3}=1$ ... (1)

• $1+\omega+\omega^{2}=0$ ... (2)

So, $1-\omega-\omega^{2}$

$=1-(\omega+\omega^{2})$

$=1-(-1)$ [by (2)]

$=1+1$

$=2$

$\Rightarrow \boxed{1-\omega-\omega^{2}=2}$ ... (3)

and $1-\omega+\omega^{2}$

$=(1+\omega^{2})-\omega$

$=(-\omega)-\omega$ [by (2)]

$=-\omega-\omega$

$=-2\omega$

$\Rightarrow \boxed{1-\omega+\omega^{2}=-2\omega}$ ... (4)

Now, $(1-\omega-\omega^{2})^{3}+(1-\omega+\omega^{2})^{3}$

$=(2)^{3}+(-2\omega)^{3}$ [using the above relations in (3) and (4)]

$=8-8\omega^{3}$

$=8-8(1)\quad[\because\omega^{3}=1]$

$=8-8$

$=\bold{0}$