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
 A: 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.
A: Addendum added where I change my mind about what the problem composer intends.  Leaving my original response undeleted for clarity, because the wording of the problem is confusing.

I consider the answer posted by mathcounterexamples.net incomplete or wrong, depending on your point of view.  The constraints of the problem are impossible to meet, because it is impossible for $a$ to be between two of the roots.
Assuming that I have made no analytical error, this implies that the book solution presented in this re-posting of the question is also wrong.
I will adopt the syntax already used in the answer of mathcounterexamples.net.  I definitely agree with the analysis in the first part of his answer.
So:

*

*the roots are given by $k/r, k, kr ~: k \neq 0 \neq r.$

*$k = 1/2$.

*$1/r+1+r = a/4 \implies 4(1/r + 1 + r) = a$.

*It is required that $a$ falls within the range of the $3$ roots.
I will show that this is impossible.

I will consider the $4$ cases of $r \geq 1, 0 < r < 1, -1 < r < 0,$ and $r \leq -1$ separately.

$\underline{\text{Case 1:} ~r \geq 1}$
The largest root is $(r/2)$.
$a > 4r$. 
Therefore, $a$ can not be between two of the roots.

$\underline{\text{Case 2:} ~0 < r < 1}$
The largest root is $(1/2r)$.
$a > 4/r$. 
Therefore, $a$ can not be between two of the roots.

$\underline{\text{Case 3:} ~-1 < r < 0}$
The smallest root is $(1/2r).$ 
Consider the constraint $~a \geq (1/2r) \implies $ 
$(1/2r) \leq 4(1/r + 1 + r) \implies $ 
$(1/2) \geq 4 + 4r + 4r^2 = (2r + 1)^2 + 3.$
This is impossible.

$\underline{\text{Case 4:} ~r \leq -1}$
The smallest root is $(r/2).$ 
Consider the constraint $~a \geq (r/2) \implies $ 
$(r/2) \leq 4(1/r + 1 + r) \implies $ 
$0 \leq 4/r + 4 + 7r/2 \implies $ 
$0 \geq 4 + 4r + 7r^2/2 \implies $ 
$0 \geq 56 + 56r + 49r^2 = (7r + 4)^2 + 40.$
This is impossible.

Addendum 
Reacting to the subsequent comment of dxiv.
He is suggesting that the problem composer never intended that $\lambda_1, \lambda_2$ refer to $2$ of the $3$ real roots of the equation.  His suggestion never occurred to me.
In retrospect, I think that he is right.
I am leaving the answer undeleted for clarity, and because, as dxiv indicated, the wording of the problem is confusing.  Assuming that dxiv is right, then I can no longer regard the answer of mathcounterexamples.net as incomplete or wrong.
