Solving $D=\begin{vmatrix}a&\omega b&\omega^2c\\\omega^2b&c&\omega a\\\omega c&\omega^2a&b\end{vmatrix}$ 
If $D=\begin{vmatrix}a&\omega b&\omega^2c\\\omega^2b&c&\omega a\\\omega c&\omega^2a&b\end{vmatrix}$ and $D'=\begin{vmatrix}a& b&c\\b&c& a\\\ c&a&b\end{vmatrix}$, where $\omega$ is the non real cube root of unity, then which of the following does not hold:



*

*$1)\ D=0$ if $a+b+c=0$ and $a,b,c$ all distinct.

*$2)\ D'=0$ if $a=b=c$ and $a+b+c\ne0$

*$3)\ D=-D'$

*$4)\ D=D'$

My Attempt:
I am getting $D'=-\frac12(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$. So, $2)$ holds.
For $D$, I tried the row operation $R_1\to R_1+R_2+R_3$ but couldn't conclude.
I also tried multiplying each row with $\omega$ twice but in vain.
I tried writing the value of $\omega$ in $i$ but couldn't finish again.
Any help?
 A: Using the Rule of Sarrus we obtain, with $\omega^3=1$,
$$
\det (D)=\det(D')= - a^3 + 3abc - b^3 - c^3.
$$
A: We know that multiplying a row of a determinant by a constant changes the determinant by that constant. Therefore,
$$ \begin{vmatrix} a&\omega b&\omega^2c\\\omega^2b & c & \omega a\\\omega c & \omega^2 a & b \end{vmatrix} = \frac{1}{\omega^3}\begin{vmatrix} a&\omega b&\omega^2c\\b & \omega c & \omega^2 a\\ c & \omega a & \omega^2b \end{vmatrix}  $$
where we multiplied the second row by $\omega$ and third row by $\omega^2$. But actually $\omega^3=1$ so the constant factor in front is just $1$. Now, multiplying a column by a constant also has the same effect, so we multiply the second column by $\omega^2$ and third column by $\omega$ to conclude that, in fact,
$$D = \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix}.$$
So really $D=D'$. It is also easier to now compute the determinant by directly expanding it to get $D = 3abc-a^3-b^3-c^3 = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Note that the final form of the determinant is a circulant matrix, for which there is a standard formula to compute the determinant. So, this problem can be generalised to $n\times n$ matrices and you can still write out an algebraic expression for the determinant explicitly.
Also, it is generally a horrible idea to write $\omega$ out in its real and imaginary part (i.e. write $\omega = (-1+i\sqrt{3})/2$) and try to simplify from there. What you need for things to simplify is just the property that $\omega^3=1$, and you are just making your life harder by writing out the real and imaginary parts. It really won't help at all.
A: Throughout this solution, I shall be making use of the fact that $\omega^3=1$ as $\omega$ is the complex cube root of unity.
Now, consider the determinant
$$D_1=\begin{vmatrix}
1 & 0 & 0\\
0 & \omega & 0\\
0 & 0 & \omega^2
\end{vmatrix} = 1\cdot \omega \cdot \omega^2=1$$
Multiplying the matrix of $D'$ with the matrix of determinant $D_1$ we get,
$$\begin{bmatrix}
a & b\omega & c\omega^2\\
b & c\omega & a\omega^2\\
c & a\omega & b\omega^2
\end{bmatrix}$$
Taking the determinant of that matrix and using the property that $det(A\cdot B)=det(A)\cdot det(B)$ we obtain
$$D'=\begin{vmatrix}
a & b\omega & c\omega^2\\
b & c\omega & a\omega^2\\
c & a\omega & b\omega^2
\end{vmatrix}$$
I shall be multiplying with $\omega^2$ on both sides to get -
$$\omega^2\cdot D'=\omega^2\cdot\begin{vmatrix}
a & b\omega & c\omega^2\\
b & c\omega & a\omega^2\\
c & a\omega & b\omega^2
\end{vmatrix}$$
Now from the property of the determinant that any scalar factor may be multiplied across any one row or column of the determinant, I am multiplying the outside factor with the second row ($R_2$) on the RHS -
$$\omega^2\cdot D'=\begin{vmatrix}
a & b\omega & c\omega^2\\
b\omega^2 & c & a\omega\\
c & a\omega & b\omega^2
\end{vmatrix}$$
Here the property of $\omega^3=1$ was used twice. Now multiplying with $\omega$ on both sides, we get -
$$1\cdot D'=\omega\cdot\begin{vmatrix}
a & b\omega & c\omega^2\\
b\omega^2 & c & a\omega\\
c & a\omega & b\omega^2
\end{vmatrix}$$
Again I shall be multiplying the scalar factor outside with the third row of the determinant ($R_3$) -
$$ D'=\begin{vmatrix}
a & b\omega & c\omega^2\\
b\omega^2 & c & a\omega\\
c\omega & a\omega^2 & b
\end{vmatrix}$$
Again the property $\omega^3=1$ was used. If you look carefully, the determinant on the RHS is $D$. Therefore, we have -
$$\boxed{D=D'}$$
Therefore, the correct answer to the question must be (3).
This question could have been done through direct row expansion of the determinant without too much effort and you could have observed that $D=D'$.
