If the roots of the equation 6x^3-11x^2+6x-1= | vedclass

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(B) Let the roots of $6x^3-11x^2+6x-1=0$ be $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Since they are in harmonic progression,$a, b, c$ are in arithmeti

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Arithmetic Progression

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(B) Let the roots of $6x^3-11x^2+6x-1=0$ be $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Since they are in harmonic progression,$a, b, c$ are in arithmetic progression.Substituting $x = \frac{1}{y}$ in the given equation,we get $6(\frac{1}{y})^3 - 11(\frac{1}{y})^2 + 6(\frac{1}{y}) - 1 = 0$,which simplifies to $y^3 - 6y^2 + 11y - 6 = 0$.Now,consider the equation $x^3-6x^2+11x-6=0$.Since $x=1$ satisfies the equation $(1-6+11-6=0)$,$(x-1)$ is a factor.Dividing by $(x-1)$,we get $(x-1)(x^2-5x+6)=0$,which factors to $(x-1)(x-2)(x-3)=0$.The roots are $1, 2, 3$.Since $2-1 = 1$ and $3-2 = 1$,the roots are in Arithmetic Progression.

If the roots of the equation $6x^3-11x^2+6x-1=0$ are in harmonic progression,then the roots of $x^3-6x^2+11x-6=0$ will be in

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