Question

which among the following is a root of the equation $x^6-2x^3+2=0$ is :

• A. $2^{1/6}{e}^{\pi i/12}$
• B. $2^{1/2}{e}^{\pi i/12}$
• C. $6e^{\pi i/12}$
• D. $e^{\pi i/12}$

Answer 1

The correct option is A $2^{1/6}{e}^{\pi i/12}$

Let $x^3=z$
Hence the equation we get $z^2-2z+2=0$
$z=x^3=1+i\text{or}z=x^3=1-i$
By using formula, we get
$x^3=\sqrt{2}\left(\frac{1+i}{\sqrt{2}}\right)$
$x=2^{\frac{1}{6}}{e}^{i\frac{(2k\pi +\frac{\pi }{4})}{3}},k\in 0,1,2$

Hence roots of the $x^3=1+i$ is
$2^{\frac{1}{6}}{e}^{i\frac{\pi }{12}},2^{\frac{1}{6}}{e}^{i\frac{9\pi }{12}},2^{\frac{1}{6}}{e}^{i\frac{17\pi }{12}}$
Similarly for $x^3=1-i$
Here $\theta =-\frac{\pi }{4}$
and roots are
$2^{\frac{1}{6}}{e}^{-i\frac{\pi }{12}},2^{\frac{1}{6}}{e}^{i\frac{7\pi }{12}},2^{\frac{1}{6}}{e}^{i\frac{15\pi }{12}}$
Hence $6$ roots of the equation are
$2^{\frac{1}{6}}{e}^{i\frac{\pi }{12}},2^{\frac{1}{6}}{e}^{i\frac{9\pi }{12}},2^{\frac{1}{6}}{e}^{i\frac{17\pi }{12}},2^{\frac{1}{6}}{e}^{-i\frac{\pi }{12}},2^{\frac{1}{6}}{e}^{i\frac{7\pi }{12}},2^{\frac{1}{6}}{e}^{i\frac{15\pi }{12}}$