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Evaluate the roots of

$$x^3-6x-6=0$$

I solved it using Cardano's method, but I'm looking for other elementary approaches through substitutions and properties of polynomials.

Thanks.

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  • $\begingroup$ Are you aware the equation is irreducible? So, the roots cannot be expressed more simply than with Cardano's method, so that is your best bet. $\endgroup$ – Oscar Lanzi Apr 21 '16 at 10:28
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$\text{Let } x = z + \frac{2}{z}\\\begin{align*}\left(z+\frac{2}{z}\right)^3 - 6\left(z+\frac{2}{z}\right) - 6 &= 0\\z^6 - 6z^3 + 8 &= 0\\z^3 &= 3\pm1\, (\text{using the quadratic formula})\\z^3 &= 4, 2,\\z &= \sqrt[3]{2},\sqrt[3]{4}\end{align*}$

Now just substitute back one value for $z$ to get one root: $x = \sqrt[3]{2} + 2\sqrt[3]{4}$

The rest of the roots can now be gotten through long division, or by using the imaginary cube roots of $2$ or $4$.

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  • $\begingroup$ Why did you set $x= z + \dfrac{2}{z}$ ? $\endgroup$ – Henry Apr 21 '16 at 10:43
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    $\begingroup$ Making the substitution $x = z + \frac{s}{z}$ turns the equation into $z^3 + z(3s-6)+ (3s^2 - 6s)\frac{1}{z} + \frac{s^3}{z^3} - 6 = 0$. Note that we can make the coefficient of the $z$ and the $\frac{1}{z}$ terms 0 by letting $s = 2$, giving us a quadratic in $z^3$ that we can easily solve. $\endgroup$ – KoA Apr 21 '16 at 10:48
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    $\begingroup$ In general we can find the roots of any depressed cubic (a cubic of the form $x^3 + ax + b$) by making this type of subsitution. $\endgroup$ – KoA Apr 21 '16 at 10:50
  • $\begingroup$ Nice! Thanks a lot. $\endgroup$ – Henry Apr 22 '16 at 16:01
  • $\begingroup$ This is Viète's trick, but it's essentially the same as Cardano's formula. $\endgroup$ – egreg Apr 22 '16 at 16:55
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Hint: $x^3-6x-6=x^3-3bcx+b^3+c^3$

$$bc=-2, b^3+c^3=-6$$ $$b=-\sqrt[3]{2}, c=-\sqrt[3]{4}$$

So $(b+c)^3=b^3+c^3+3bc(b+c)$, then number $b+c$ - solution of $x^3-3bcx+b^3+c^3=0$

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  • $\begingroup$ How did you set the coefficients $b, c$ then? $\endgroup$ – Colescu Apr 22 '16 at 14:15
  • $\begingroup$ @Yuxiao Xie: $c=-\frac{2}{b}$ $$b^3-\frac{8}{b^3}=-6$$ $$b^6+6b^3-8=0$$ $$b^3=t$$ $$t^2+6t-8=0$$ $\endgroup$ – Roman83 Apr 22 '16 at 15:04
  • $\begingroup$ I mean, how did you know it should be something like $x^3 - 3bcx + b^3 + c^3$? $\endgroup$ – Colescu Apr 22 '16 at 15:05
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    $\begingroup$ ursoswald.ch/download/CUBIC.pdf $\endgroup$ – Roman83 Apr 22 '16 at 15:10

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