polynomials!

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}