Question

Let x be a nonreal complex number satisfying $(x-1{)}^3+8=0$

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

Answer 1

Answer: (B), (C)

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

$(x-1{)}^3+8=0$
OR
$(X{)}^3+(2.1{)}^3=0$
This is similar to the expression
$x^3+1=0$
whose roots are $(-1,-w,-w^2)$
Hence
$X=(-2,-2w,-2w^2)$
$(x-1)=(-2,-2w,-2w^2)$
$x=-1$, $x=1-2w$, $x=1-2w^2$