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

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Solution

Put $x = 1$
$1 - 7 + 14 - 8 = 0$
So $x - 1$ is a factor of given equation
Dividing given equation by $(x - 1)$ we get
$x^{2} - 6 x + 8$
So the given equation is
$(x - 1) (x^{2} - 6 x + 8) = 0$
$(x - 1) (x - 2) (x - 4) = 0$
$x = 1, 2, 4$

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