If the roots of the equation $x^3-3x^2+4x-1=0$ are $\alpha ,\beta ,$ and $\gamma$, then find the value of $({\alpha }^2+{\beta }^2+{\gamma }^2)$.

Solution

1. For the cubic equation $x^3-3x^2+4x-1=0$, Vieta's formulas give the sum of the roots $\alpha +\beta +\gamma =3$ and the sum of the roots taken two at a time $\alpha \beta +\beta \gamma +\gamma \alpha =4$.
2. Use the algebraic identity $(\alpha +\beta +\gamma {)}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\gamma \alpha )$ to express the sum of squares.
3. Substitute the known values into the identity: $3^2=({\alpha }^2+{\beta }^2+{\gamma }^2)+2(4)$.
4. Solve for the sum of squares: $9=({\alpha }^2+{\beta }^2+{\gamma }^2)+8$, which simplifies to ${\alpha }^2+{\beta }^2+{\gamma }^2=1$.
5. Hence the answer is (A).