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.
$\endgroup$ 12 Answers
$\begingroup$$\text{Let } x = z + \frac{2}{z}\\\begin{align*}\left(z+\frac{2}{z}\right)^3 - 6\left(z+\frac{2}{z}\right) - 6 &= 0\\z^6 - 6z^3 + 8 &= 0\\z^3 &= 3\pm1\, (\text{using the quadratic formula})\\z^3 &= 4, 2,\\z &= \sqrt[3]{2},\sqrt[3]{4}\end{align*}$
Now just substitute back one value for $z$ to get one root: $x = \sqrt[3]{2} + 2\sqrt[3]{4}$
The rest of the roots can now be gotten through long division, or by using the imaginary cube roots of $2$ or $4$.
$\endgroup$ 5 $\begingroup$Hint:$x^3-6x-6=x^3-3bcx+b^3+c^3$
$$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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