polynomials!

sum of the roots of the equation = 0
and as the roots can be represented as 1,a,...,a⁶
So, 1+a+a²+...+a⁶ = 0
So, a+a²+...+a⁶ = -1 .. (i)
Product of roots of eqn = 1.a.a²...a⁶ = 1 ... (ii)

Now sum of roots of reqd eqn = a+a²+...+a⁶ = -1
product of roots of reqd eqn = 2+a+a²+...+a⁶ = 1
product of roots taking 2 at a time = 2(a+a²+...+a⁶) = -2
So the reqd eqn is k(x³+x²-2x-1)=0

one can also proceed by taking x=re^iθ and then proceed

Aliter:
a, a²,..,a⁶, are roots of 1+x+x^2+x^3+...+x^6 = 0

Notice that the three given roots are of the form \alpha + \frac{1}{\alpha} where α is any root of the above equation.

Hence, if we are able to write the equation in the variable x + \frac{1}{x}, we would have found the required equation.

So, first divide by x³. So we have x^3 + \frac{1}{x^3} + x^2 + \frac{1}{x^2} + x + \frac{1}{x}+ 1 = 0

\Rightarrow \left(x + \frac{1}{x} \right)^3 - 3 \left(x + \frac{1}{x} \right) + \left(x + \frac{1}{x} \right)^2 -2 + \left(x + \frac{1}{x} \right)+1 = 0

or t³+t²-2t-1 = 0 where t = x + \frac{1}{x}