Question
Show that $\left(\frac{1+i}{\sqrt{2}}\right)^8+\left(\frac{1-i}{\sqrt{2}}\right)^8=2$
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$\left|\begin{array}{lll}x+2 & x+6 & x-1 \\ x+6 & x-1 & x+2 \\ x-1 & x+2 & x+6\end{array}\right|=0$
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get
$\left|\begin{array}{ccc}x+2 & x+6 & x-1 \\ 4 & -7 & 3 \\ -3 & -4 & 7\end{array}\right|=0$
∴ (x + 2)(- 49 + 12) – (x + 6)(28 + 9) + (x- 1)(- 16 – 21) = 0 ∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0 ∴ -37(x + 2+ x + 6 + x – 1) = 0 ∴ 3x + 7 = 0
$\therefore x=\frac{-7}{3}$