Q.Solve $x^3-7x^2+14x-8=0$, given that its roots are in geometric progression.

Roots $a-d,a,a+d$ with sum $=9/2$ give $a=3/2$; the product relation gives $d=3$, so roots are $-3/2,\,3/2,\,9/2$.

For $8x^3 - 36x^2 - 18x + 81 = 0$, let the roots in AP be $a - d,\ a,\ a + d$.

Sum of roots $= 3a = -\dfrac{-36}{8} = \dfrac{36}{8} = \dfrac{9}{2}$, so $a = \dfrac{3}{2}$.

Since $a$ is a root, substitute $x = \dfrac{3}{2}$: $8\cdot\dfrac{27}{8} - 36\cdot\dfrac{9}{4} - 18\cdot\dfrac{3}{2} + 81 = 27 - 81 - 27 + 81 = 0$. Confirmed.

Product of roots $= (a-d)\,a\,(a+d) = a(a^2 - d^2) = -\dfrac{81}{8}$.

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Roots in G.P. multiply to $8$, so the middle root is $2$; factoring gives roots $\tfrac23, 2, 6$.

Let the roots in G.P. be $\dfrac{a}{r}, a, ar$. For $3x^3 - 26x^2 + 52x - 24 = 0$, the product of the roots is

$\dfrac{a}{r}\cdot a\cdot ar = a^3 = -\dfrac{-24}{3} = 8 \;\Rightarrow\; a = 2$.

So $x = 2$ is a root. Check: $3(8) - 26(4) + 52(2) - 24 = 24 - 104 + 104 - 24 = 0$.

Divide out $(x-2)$:

$3x^3 - 26x^2 + 52x - 24 = (x-2)(3x^2 - 20x + 12)$.

Solve $3x^2 - 20x + 12 = 0$: …