If $\alpha$ and $\beta$ are roots of $(x^2)-(4x)-1=0$, find $\sqrt[3]{\alpha}+ \sqrt[3]{\beta}$ My question in handwriting
https://i.stack.imgur.com/4vPBs.jpg
If $\alpha$ and $\beta$ are roots of this equation
$$(x^2)-(4x)-1=0$$
Then find $$\sqrt[3]{\alpha}+\sqrt[3]{\beta}$$
Please do not use the $\Delta = b^2-4ac$ method. Use Vieta's formulas: $$S=\alpha+\beta=-\frac{b}{a} \qquad P=\alpha\beta=\frac{c}{a}$$
Actually I want to solve the main entry using difference of squares! Such as (x-a)(x+a)
And I note everybody that I know basical rules of quadratic equations like delta and etc.

Let $\sqrt[3]{\alpha}+\sqrt[3]{\beta}=x$.
Thus, since $\alpha+\beta=4$ and $\alpha\beta=-1,$ we obtain
$$x^3=\alpha+\beta+3\sqrt[3]{\alpha\beta}(\sqrt[3]{\alpha}+\sqrt[3]{\beta})=4+3\cdot(-1)x$$ or
$$x^3+3x-4=0$$ or
$$x^3-x^2+x^2-x+4x-4=0$$ or
$$(x-1)(x^2+x+4)=0,$$ which gives $x=1$.
