Question

If the cube roots of unity are $1$, $\omega ,{\omega }^2$, then the roots of the equation $(x-1{)}^3+8=0$ are

• A. $-1,1+2\omega ,1+2{\omega }^2$
• B. $-1,1-2\omega ,1-2{\omega }^2$
• C. $-1,-1,-1$
• D. none of these

Answer 1

Answer: (B)

$-1,1-2\omega ,1-2{\omega }^2$

Since $w$ is the cube root of unity.
$\therefore w^3=1\&1+w+w^2=0$ where $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{-1-i\sqrt{3}}{2}$
${(x-1)}^3+8=0$
$\Rightarrow x=1+2{(-1)}^{\frac{1}{3}}=1+2{(\cos \pi +i\sin \pi )}^{\frac{1}{3}}$
$\Rightarrow x=1+2(\cos \left(\frac{2k\pi +\pi }{3}\right)+i\sin \left(\frac{2k\pi +\pi }{3}\right))$ ...{ De Moivre's Theorem }
where, $k=0,1,2$
For$k=0$
$\Rightarrow x=1+2(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right))=1+2\left(\frac{1+i\sqrt{3}}{2}\right)=1-2w^2$
For $k=1$
$\Rightarrow x=1+2(\cos \pi +i\sin \pi )=1-2=-1$
For $k=2$
$\Rightarrow x=1+2(\cos \left(\frac{5\pi }{3}\right)+i\sin \left(\frac{5\pi }{3}\right))=1+2\left(\frac{1-i\sqrt{3}}{2}\right)=1-2w$
Ans: B