1. Imagenary cube root Problems

If $w$ is an imagenary cube root of unity , then find the value of

$
\begin{vmatrix}
1+w & w^2 & -w \\
1+w^2 & w & -w^2 \\
w^2+w & w & -w^2
\end{vmatrix}
$

2. did you type it right? that bottom 2x2 minor has rows the same, is that correct?

If so, just do it, i recommend cofactor down the first column to take advantage of that being the same.

the first term is 0.

second is
$-(1+w^2)[-w+w^2]$
third is
$(w^2+w)[-w+w^2]$

$(-1-w^2 +w^2 +w)[-w+w^2]=(-1+w)[-w+w^2]=1-w^2-w^2+w=-2w^2+w+1$
But these are cube roots of unity, $w^2+w=-1 \Rightarrow w=-1-w^2$

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

3. I didnt get the last part

Originally Posted by Gamma
did you type it right? that bottom 2x2 minor has rows the same, is that correct?

If so, just do it, i recommend cofactor down the first column to take advantage of that being the same.

the first term is 0.

second is
$-(1+w^2)[-w+w^2]$
third is
$(w^2+w)[-w+w^2]$

$(-1-w^2 +w^2 +w)[-w+w^2]=(-1+w)[-w+w^2]=1-w^2-w^2+w=-2w^2+w+1$
But these are cube roots of unity, $w^2+w=-1 \Rightarrow w=-1-w^2$

$-2w^2+w+1= -2w^2 (-1 - w^2) + 1= -3w^3=-3$
I didnt get the last part
$-2w^2+w+1= -2w^2 (-1 - w^2) + 1= -3w^3=-3$

could u please explain how u got this?????

4. i have some doubt

Originally Posted by zorro
I didnt get the last part
$-2w^2+w+1= -2w^2 (-1 - w^2) + 1= -3w^3=-3$

could u please explain how u got this?????
-Firstly the $-3w^3$ shouldnt it be $-3w^2$

-Secondly where did the $w^2$ go and how did $-3w^2 \ = \ -3$

-Firstly the $-3w^3$ shouldnt it be $-3w^2$
-Secondly where did the $w^2$ go and how did $-3w^2 \ = \ -3$