Explanation

(a)
Since, the roots of the equation $8x^{3} - 14x^{2} + 7x - 1 = 0$ are in GP. Let the roots be $\frac{\alpha}{\beta}$,$\ \alpha,\ \alpha\beta,\ \beta \neq 0$. Then, the product of roots is $\alpha^{3} = \frac{1}{8} \Rightarrow \alpha = \frac{1}{2}\ and\ hence,\ \beta = \frac{1}{2}$.
So, roots are 1, $\frac{1}{2},\frac{1}{4}$.