Solve the equation `root(3)((2x-1))+root(3)((x-1))=1`

Explanation

We have `root(3)((2x-1))+root(3)((x_1))=1`……I
Cubing both sides of Eq. (i) we obtain
`2x-1+x-1s+3.root(3)((2x-1)(x-1))`
`(root(3)((2x-1))+root(3)((x-1)))=1`
`implies3x-2+3.root(3)((2x^(2)-3x+1))(1)=1`[from Eq. (i) ]
`implies3.root(3)((2x^(2)-3x=1))=3-3x`
`impliesroot(3)((2x^(2)-3x+1))=(1-x)`
Again cubing both sides, we obtain
`2x^(2)-3x+1=(1-x)^(3)`
`implies(2x-1)(x-1)=(1-x)^(3)`
`implies(2x-1)(x-1)=-(x-1)^(3)`
`implies(x-1){2x-1+(x-1)^(2)}=0`
`implies(x-1)(x^(2))=0`
`:.x_(1)=0` and `x_(2)=1`
`:'x_(1)=0` is not satisfied the Eq. (i) then `x_(1)=0` is an extraneous root of the eq. (i) thus` x_(2)=1` is the only root of the original equation.