Mathematics - Quadratic Equation Question with Solution | TestHub

For the equation $x^2-3x+1=0$, we have $\alpha +\beta =-(-3)/1=3$ and $\alpha \beta =1/1=1$. We need to find the value of ${\alpha }^3+{\beta }^3$. We use the identity ${\alpha }^3+{\beta }^3=(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)$. This can be further written as ${\alpha }^3+{\beta }^3=(\alpha +\beta )[(\alpha +\beta {)}^2-3\alpha \beta ]$. Substituting the values of $\alpha +\beta$ and $\alpha \beta$: ${\alpha }^3+{\beta }^3=(3)[(3{)}^2-3(1)]=3[9-3]=3(6)=18$.