Discussion Forum

 If x²+x+1= 0 , then the value of   $\sum_{n=1}^{6}\left(x^{n}+\frac{1}{x^{n}}\right)^{2}$  is

Answer:

Explanation:

 Gtiven equation is x²+x+1=0

 $\Rightarrow x=\omega$ and $x=\omega^{2}$

 Case I ; when x=$\omega$

 Then

$\sum_{n=1}^{6}\left[x^{n}+\frac{1}{x^{n}}\right]^{2}=\sum_{n=1}^{6}\left[\omega^{n}+\omega^{2n}\right]^{2}\left[\because \frac{1}{\omega}=\omega^{2}\right]$

$=(\omega+\omega^{2})^{2}+(\omega^{2}+\omega^{4})^{2}+(\omega^{3}+\omega^{6)^{2}}+(\omega^{4}+\omega^{8)^{2}}+(\omega^{5}+\omega^{10)^{2}}+(\omega^{6}+\omega^{12)^{2}}$

$=(-1)^{2}+(-1)^{2}+(2)^{2}+(-1)^{2}+(-1)^{2}+(2)^{2}=12$

 Case II : when $\omega^{2}$

 Then

  $\sum_{n=1}^{6}\left[x^{n}+\frac{1}{x^{n}}\right]^{2}=\sum_{n=1}^{6}\left[\omega^{2n}+\omega^{n}\right]^{2}\left[\because\frac{1}{\omega^{2}}=\omega\right]$

=12