Each of the roots of the equation $x^3 - 6x^2 + 6x-5 =0$ are increased by $h$. So that the new transformed equation does not contain $x^2$ term, then $h$ is equal to :

Question

( A ) $2$
( B ) $\frac{1}{3}$
( C ) $\frac{1}{2}$
( D ) $1$