Question

Solution set of $x^6-x^5+x^4-x^2+x-1=0$, $xϵ0$ is

• A. $\{\pm 1,\pm i,\pm \sqrt{2}\}$
• B. $\{0,\pm 1,\pm i\}$
• C. $\{\pm 1,\pm i,\frac{1}{2}(1\pm i\sqrt{3})\}$
• D. $\{\pm 1,\pm i\}$

Answer 1

Answer: (C)

$\{\pm 1,\pm i,\frac{1}{2}(1\pm i\sqrt{3})\}$

$x^6-x^5+x^4-x^2+x-1=0$
$x^4(x^2-x+1)-1(x^2-x+1)=0$
$(x^4-1)(x^2-x+1)=0$
$x^4=1$
$x^2=\pm 1$
$x^2=1$
$x=\pm 1$ and $x^2=-1$
$x=\pm i$
Now
$x^2-x+1=0$
$x=\frac{1\pm i\sqrt{(}3)}{2}$
Hence the answer is Option C.