Question

The values of $x$ which satisfies the equation $(x-1{)}^3+8=0$ are

• A. $-1,1-2\omega \text{and}1-2{\omega }^2$
• B. $1,1+2\omega \text{and}1+2{\omega }^2$
• C. $-1,1+2\omega \text{and}1+2{\omega }^2$
• D. $1,1-2\omega \text{and}1-2{\omega }^2$

Answer 1

The correct option is A $-1,1-2\omega \text{and}1-2{\omega }^2$

We have,
$(x-1{)}^3+8=0\Rightarrow (x-1{)}^3=-8\Rightarrow x-1=(-8{)}^{1/3}$
$\Rightarrow x-1=2(-1{)}^{1/3}$
$\Rightarrow x-1=2(-1)\text{or}x-1=2(-\omega ),x-1=2(-{\omega }^2)$
$[\because (-1{)}^{1/3}=-1\text{or}-\omega \text{or}-{\omega }^2]$
$\Rightarrow x-1=-2\text{or}x-1=-2\omega \text{or}x-1=-2{\omega }^2$
$\Rightarrow x=-1\text{or}x=1-2\omega \text{or}x=1-2{\omega }^2$.
Hence, the solutions of the equation $(x-1{)}^3+8=0\text{are}-1,1-2\omega \text{and}1-2{\omega }^2$