Question

if w is a complex cube root of unity show that (1+w)³-(1+w²)³=0​

Answer 1

ω is a complex cube root of unity.

$\omega ^3=1$

→$\omega^3-1=0$

On solving with identity,

the factors of $\omega ^3-1$ are $\omega ^2+\omega +1$ and $\omega - 1$.

Since $D=b^2-4ac$ of $\omega ^2+\omega +1$ is negative,

we can observe ω is a solution of $\omega ^2+\omega +1=0$.

Now,

$(1+\omega)^3-(1+\omega ^2)^3$

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

$=(1+3\omega+3\omega^2+1)-(1+3\omega^2+3\omega\times 1+1^2)$

$=2+3\omega +3\omega^2-(2+3\omega^2+3\omega)$

$=(2-2)+(3\omega-3\omega)+(3\omega^2-3\omega^2)$

$=0$

Hence shown!

To solve, use

• $\omega^3=1$
• $\omega^2+\omega+1=0$

Answer 2

the answer of above question is 0(Zero)