If $2,1,1$ are roots of the equation $x^3-4x^2+5x-2=0$

Putting $2, 1, 1$ in the equation we get:

$2^3+4\times 2^2+5\times 2-2=0;$

$1^3+4\times 1^2+5\times 1-2=0;$

$1^3+4\times 1^2+5\times 1-2=0;$

To find the roots of second equation we will subtract $\frac{1}{3}$ from the roots of the first equation,

So, the roots are $2-\frac{1}{3}, 1-\frac{1}{3}, 1-\frac{1}{3}\equiv \frac{5}{3},\frac{2}{3},\frac{2}{3}$

${\left(x+\frac{1}{3}\right)}^3$ $-4{\left(x+\frac{1}{3}\right)}^2+5\left(x+\frac{1}{3}\right)-2=0$

Putting $\frac{5}{3},\frac{2}{3},\frac{2}{3}$ in the equation we get

$2^3+4\times 2^2+5\times 2-2=0;$

$1^3+4\times 1^2+5\times 1-2=0;$

$1^3+4\times 1^2+5\times 1-2=0;$

Which satisfies the equation.

Hence, $\frac{5}{3},\frac{2}{3},\frac{2}{3}$ are the roots of the new equation.