Could anyone help me to find how to find all solution for $x^3=6x+6$?
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Try http://mathworld.wolfram.com/CubicFormula.html. – Oscar Lanzi Feb 24 '17 at 11:09
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1See http://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0/2157645#2157645 – lab bhattacharjee Feb 24 '17 at 11:10
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By Eisenstein's Criterion with $p = 3$, no ration roots exists, so you really have to use the formula in the above link. – GNUSupporter 8964民主女神 地下教會 Feb 24 '17 at 11:10
2 Answers
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Use Cardano's method:
Set $y=u+v$. As you have two unknowns instead of one, you can add a constraint on $u,v$, in order to simplify the equation. This equation becomes $$(u+v)^3-6(u+v)-6=u^3+v^3+(u+v)(3uv-6)-6=0.$$ Adding the condition $3uv=6$, i.e. $uv=2$, you obtain the equation $u^3+v^3-6=0$, whence the system $$\begin{cases}u^3+v^3=6\\u^3v^3=8\end{cases}\iff u^3, v^3\;\text{ are the roots of the quadratic equation}\quad t^2-6t+8=0.$$ Now $t^2-6t+8=(t-3)^2-9+8$, so $u^3, v^3=3\pm1=2,4$ and finally $$u,v=\sqrt[3]{2},\; \sqrt[3]{4},\enspace\text{whence}\quad y=\sqrt[3]{2}+\sqrt[3]{4}.$$
Bernard
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Set $x=ay$; the equation becomes $a^3y^3-6ay=6$. We want that $$ \frac{a^3}{6a}=\frac{4}{3} $$ so we can take $a=2\sqrt{2}$. Then we have $$ 16\sqrt{2}y^3-12\sqrt{2}y=6 $$ that becomes $$ 4y^3-3y=\frac{3}{4}\sqrt{2}>1 $$ Now set $y=\cosh z$, so the equation becomes $$ 4\cosh^3z-3\cosh z=\frac{3}{4}\sqrt{2} $$ that is, $$ \cosh3z=\frac{3}{4}\sqrt{2} $$ Now solve $$ e^{3z}+e^{-3z}=\frac{3}{2}\sqrt{2} $$ and you'll have $z$ and so $y$ and finally $x$.
egreg
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