Q.If $\omega\ne1$ is a cubic root of unity and $\begin{vmatrix}1&1&1\\1&-\omega^2-1&\omega^2\\1&\omega^2&\omega^7\end{vmatrix}=3k$, then $k$ is equal to

Lists the four values of the fractional power by de Moivre's theorem and multiplies them, using that the sum of their arguments is a multiple of $2\pi$.

1. Write $z=\cos\dfrac\pi3+i\sin\dfrac\pi3=e^{i\pi/3}$.
2. By the generalised de Moivre theorem, the four values of $z^{3/4}$ are $w_k=\cos\left(\dfrac{3\left(\frac\pi3+2k\pi\right)}4\right)+i\sin\left(\dfrac{3\left(\frac\pi3+2k\pi\right)}4\right)$ for $k=0,1,2,3$, i.e. arguments $\theta_k=\dfrac{\pi+6k\pi}4$.
3. $\theta_0=\dfrac\pi4$, $\theta_1=\dfrac{7\pi}4$, $\theta_2=\dfrac{13\pi}4$, $\theta_3=\dfrac{19\pi}4$. …

Writing the square root as $x+iy$ and equating real and imaginary parts of $(x+iy)^2=i$ pins down $x=y=1/\sqrt2$.

1. Let $\sqrt i=x+iy$ for real $x,y$. Then $(x+iy)^2=i$.
2. Expand: $x^2-y^2+2xyi=0+1\cdot i$.
3. Equate real parts: $x^2-y^2=0\Rightarrow x^2=y^2\Rightarrow x=\pm y$.
4. Equate imaginary parts: $2xy=1$, so $xy>0$, meaning $x$ and $y$ have the same sign — hence $x=y$ (not $x=-y$). …

Rewriting each bracket using $1+\omega+\omega^2=0$ as $-2\omega$ and $-2\omega^2$, then simplifying powers with $\omega^3=1$, gives the value $-16$.

1. Recall the key property of a cube root of unity $\omega\neq1$: $1+\omega+\omega^2=0$ and $\omega^3=1$.
2. Rewrite $1-\omega+\omega^2 = (1+\omega+\omega^2) - 2\omega = 0-2\omega = -2\omega$.
3. Rewrite $1+\omega-\omega^2 = (1+\omega+\omega^2) - 2\omega^2 = 0-2\omega^2 = -2\omega^2$.
4. So the expression becomes $(-2\omega)^4 + (-2\omega^2)^4 = 16\omega^4 + 16\omega^8$. …

Reduce all powers of $\omega$ mod 3, use $(1-\omega)(1-\omega^2)=3$ (a standard cube-root-of-unity identity), then square it.

1. $\omega$ is a complex cube root of unity, so $\omega^3=1$ and $1+\omega+\omega^2=0$.
2. Reduce the exponents modulo 3: $\omega^4=\omega^{3+1}=\omega$, and $\omega^8=\omega^{6+2}=\omega^2$.
3. So the product becomes $(1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2) = \big[(1-\omega)(1-\omega^2)\big]^2$. …

The expression equals $\omega+\omega^2=-1$, where $\omega$ is a complex cube root of unity.

1. Let $\omega=\dfrac{-1+i\sqrt3}{2}$; then $\omega^2=\dfrac{-1-i\sqrt3}{2}$, with $\omega^3=1$ and $1+\omega+\omega^2=0$.
2. $\omega^{100}=\omega^{99}\cdot\omega=(\omega^3)^{33}\cdot\omega=1^{33}\cdot\omega=\omega$.
3. $(\omega^2)^{100}=\omega^{200}=\omega^{198}\cdot\omega^2=(\omega^3)^{66}\cdot\omega^2=\omega^2$. …