Solve `6x^(3)-11x^(2)+6x-1=0`, roots of the equatioin are in HP.

Explanation

Put `x=1/y` in the given equation then
`6/(y^(3))-11/(y^(2))+6/y-1=0`
`impliesy^(3)-6y^(2)+11y-6=0`……..i
Now, roots of Eq. (i) are in AP.
Let the roots be `alpha-beta, alpha, alpha +beta`
The, sum of roots `=alpha-beta+alpha+beta=6`
`implies3 alpha=6`
`:.alpha=2`
Product of roots `=(alpha-beta).alpha.(alpha+beta)=6`
`implies(2-beta)2(2+beta)=6implies4-beta^(2)=3`
`:.beta=+-1`
`:.` Roots of Eqs (i) are 1,2,3 or 3,2,1.
Hence roots of the given equation are `1,1/2,1/3` or `1/3, 1/2, 1`.