If there is a multiple root of order 3 for the equation ${\mathrm{x}}^4 – 2{\mathrm{x}}^3 + 2\mathrm{x} – \mathrm{a} = 0$ then the other root is

$\begin{array}{l}Let\alpha ,\alpha ,\alpha ,\beta arerootsof{x}^4-2x^3+2x-a\\ \therefore \alpha +\alpha +\alpha +\beta =2\\ \Rightarrow 3\alpha +\beta =2--\left(1\right)\\ \therefore {\alpha }^2+{\alpha }^2+{\alpha }^2+\alpha \beta +\alpha \beta +\alpha \beta =0\\ \Rightarrow 3{\alpha }^2+3\alpha \beta =0\\ \Rightarrow 3\alpha (\alpha +\beta )=0[\because \alpha \ne 0]\\ \Rightarrow \alpha +\beta =0\\ \Rightarrow \alpha =-\beta \\ from\left(1\right)\\ \Rightarrow -3\beta +\beta =2\\ \Rightarrow -2\beta =2\\ \Rightarrow \beta =-1\end{array}$