If there is a multiple root of order 3 for the equation $x^4-2x^3+2x-a=0$ , then the other root is

$f^1\left(x\right)=4x^3-6x^2+2=0$

$2x^3-3x^2+1=0$

$\left(x-1\right)\left(2x^2-2x+x-1\right)=0$

$\left(x-1\right)\left(2x\left(x-1\right)+1\left(x-1\right)\right)=0$

$x=1,1,\frac{-1}{2}$

1 is root of multiplicity 3

$1+1+1+\delta =2$

$\therefore \delta =-1$