Given, $\sum _{k=1}^n\left(k+\frac{1}{\omega }\right)\left(k+\frac{1}{{\omega }^2}\right)=340$

$\sum _{k=1}^n\left(1+\frac{1}{\omega }\right)\left(1+\frac{1}{{\omega }^2}\right)+\left(2+\frac{1}{\omega }\right)\left(2+\frac{1}{{\omega }^2}\right)+........+\left(n+\frac{1}{\omega }\right)\left(n+\frac{1}{{\omega }^2}\right)=340$

Now, $\sum _{k=1}^n\left(k+\frac{1}{\omega }\right)\left(k+\frac{1}{{\omega }^2}\right)=340$

$\sum _{k=1}^n\left(k^2+\frac{k}{\omega }+\frac{k}{{\omega }^2}+\frac{1}{{\omega }^3}\right)=340$

$\sum _{k=1}^n\left[k^2+k\left(\frac{1}{\omega }+\frac{1}{{\omega }^2}\right)+\frac{1}{{\omega }^3}\right]=340$

$\sum _{k=1}^n\left[k^2+k\left(\frac{\omega +1}{{\omega }^2}\right)+\frac{1}{{\omega }^3}\right]=340$

We know that  ${\omega }^3=1, 1+\omega +{\omega }^2=1$

$\sum _{k=1}^n\left[k^2+k\left(\frac{-{\omega }^2}{{\omega }^2}\right)+1\right]=340$

$\sum _{k=1}^n\left[k^2-k+1\right]=340$

Now in terms of n

$\underset{k=1}{\overset{n}{\sum }}{n}^2-\underset{k=1}{\overset{n}{\sum }}n+\underset{k=1}{\overset{n}{\sum }}1=340$

$\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}+n=340$

$\frac{n\left(n+1\right)}{2}\left[\frac{\left(2n+1\right)}{3}-1\right]+n=340$

$\frac{n\left(n+1\right)}{2}\left[\frac{2\left(n-1\right)}{3}\right]+n=340$

$\frac{n\left(n+1\right)\left(n-1\right)}{3}+n=340$

$n^3+2n-1020=0$

Putting $n=10$

$1020-1020=0$

$\left(n-10\right)$is factor of equation $n^3+2n-1020=0$

Equation $n^3+2n-1020=0$ Dividing by $\left(n-10\right)$

Then, $\left(n-10\right)\left(n^2+10n+102\right)=0$

If $n-10=0$

$n=10$

If  $n^2+10n+102=0$

$a=1, b=10, c=102$

$x=\frac{-10\pm \sqrt{100-408}}{2}$

$x=\frac{-10\pm \sqrt{-308}}{2}$

Value of $x$ is not possible, because value of $x$ is complex

Hence, value of $x=10$