# 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 $$x=1$$.
• We get that $$\sqrt[3]{\alpha}+\sqrt[3]{\beta}$$ are NOT real numbers Oct 11, 2018 at 9:55
• Ok this is an old discussion about $$\sqrt[3]{x}$$! Oct 11, 2018 at 10:08