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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.

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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 $x=1$.

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  • $\begingroup$ We get that $$\sqrt[3]{\alpha}+\sqrt[3]{\beta}$$ are NOT real numbers $\endgroup$
    – Dr. Sonnhard Graubner
    Oct 11, 2018 at 9:55
  • $\begingroup$ Why do you think so, Sonnhard? By the way, how are you? $\endgroup$
    – Michael Rozenberg
    Oct 11, 2018 at 9:56
  • $\begingroup$ I'm here in Austria for a few days $\endgroup$
    – Dr. Sonnhard Graubner
    Oct 11, 2018 at 9:59
  • $\begingroup$ Beautiful country! $\endgroup$
    – Michael Rozenberg
    Oct 11, 2018 at 10:00
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    $\begingroup$ Ok this is an old discussion about $$\sqrt[3]{x}$$! $\endgroup$
    – Dr. Sonnhard Graubner
    Oct 11, 2018 at 10:08

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