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Question

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


A
{±1,±i,±2}
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B
{0,±1,±i}
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C
{±1,±i,12(1±i3)}
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D
{±1,±i}
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Solution

The correct option is B $$\left \{ \pm 1,\pm i,\dfrac{1}{2}\left ( 1\pm i\sqrt{3} \right ) \right \}$$
$$x^{6}-x^{5}+x^{4}-x^{2}+x-1=0$$
$$x^{4}(x^{2}-x+1)-1(x^{2}-x+1)=0$$
$$(x^{4}-1)(x^{2}-x+1)=0$$
$$x^{4}=1$$
$$x^{2}=\pm1$$
$$x^{2}=1$$
$$x=\pm1$$ and $$x^{2}=-1$$
$$x=\pm i$$
Now
$$x^{2}-x+1=0$$
$$x=\dfrac{1\pm i \sqrt(3)}{2}$$
Hence the answer is Option C.

Maths

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