Question

The value of ${\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^6+{\left(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\right)}^6$ is

• A. $-2$
• B. $0$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Replacing the above terms by cube root of unit $w$ we get
$(\frac{w^2}{w}{)}^6+(\frac{w}{w^2}{)}^6$
$=w^6+\frac{1}{w^6}$
$=(w^3{)}^2+\frac{1}{(w^3{)}^2}$
$=1+1$
$=2$