Three real roots of $8x^3 – ax^2 + bx – 1 = 0$ in G.P. The equation $8x^3 – ax^2 + bx – 1 = 0$ has three real roots in G.P. If $λ_1 ≤ a ≤ λ_2$, then find ordered pair $(λ_1, λ_2)$.
My approach $f(x)=8x^3 – ax^2 + bx – 1 $
$f'(x)=24x^2 –2ax + b$
For real root 
${(4a^2-96b)}>0$ 
Roots of $f'(x)$ are
$T=\frac{2a+\sqrt{(4a^2-96b)}}{48}$ &U= $\frac{2a-\sqrt{(4a^2-96b)}}{48}$
Now $f(T).f(U)<0$ then we have three real roots
If above condition is satisfied then 
we need to frame another equation
Let the roots are $a',a'r,a'r^2$
$a'+a'r+a'r^2=\frac{a}{8}$
$a'^2(r+r^2+r^3)=\frac{b}{8}$
$a'^3r^3=\frac{1}{8}$
from here I am not able to proceed

Suppose that the 3 roots are $k/r, k, kr$. Their product $k^3$ is equal to $1/8$. Therefore, $k = 1/2$.
Their sum is $a/8 = k(1/r+1+r)$. Hence $1/r+1+r = a/4$.
Finally, $k^2(1/r+1+r)= b/8$, which implies $(1/r+1+r)=b/2$.
If a solution exists, we must have $a =2b$ and $r^2 + (1-b/2) r+1=0$. If this last equation has real roots, its discriminant has to be positive
$$\Delta = (1-b/2)^2-4 \ge 0$$ which means 
$$1-b/2 \in (-\infty , -2] \cup [2, \infty)$$ or
$$b \in (-\infty ,2] \cup [6, \infty)$$ with $a =2b$.
Conversely, if those conditions are met, the roots will form a geometric progression.
