In my course of Galois Theory I have this problem, I have to prove that: $$\sqrt[3]{-18+\sqrt{325}} + \sqrt[3]{-18-\sqrt{325}} =3 $$
I have already tried to equal it to $x$ And expand $x^3$ but after all the process the result is that $$ x^3+3x+36=0$$ and I need $$x^3+3x-36=0$$
Please some help with this.
On
You have: $x^3+y^3 = -36, xy = -1$. Thus: $(x+y)^3 = x^3+3xy(x+y) + y^3 \implies s^3 = -3s - 36\implies s^3+3s+36 = 0\implies (s+3)(s^2-3s + 12) = 0\implies s = -3$ since $s^2-3s+12 = \left(s-\dfrac{3}{2}\right)^2+ \dfrac{39}{4} > 0$.
Note: the plus and minus signs of $3$ is not significantly important here although the correct answer is wanted. Point is find a factorable polynomial equation of $s = x+y$.
On
I solved this kind of problem before and noticed we can actually find a closed form for the cubic roots in some cases:
Your cubic roots are of the form $\sqrt[3]{-18\pm 5\sqrt{13}}$
According to my previous study, we can try to search for real roots of the type $\dfrac{x\pm y\sqrt{13}}2$
Such that the sum of the two roots is $x$.
Since here you are guided about $x=-3$ then let search for $y$.
$\left(\dfrac{-3\pm y\sqrt{13}}2\right)^3=\frac 18(-27-117y^2)\pm\frac 18(27y+13y^3)\sqrt{13}$
$\begin{cases} -27-117y^2=-18\times 8\\ 27y+13y^3=5\times 8\end{cases}\implies y=1$
Thus $\sqrt[3]{-18\pm 5\sqrt{13}}=\dfrac{-3\pm \sqrt{13}}2$
Let $\alpha=\sqrt[3]{\sqrt{325}+18}$ and $\beta=\sqrt[3]{\sqrt{325}-18}$. Since $18^2=324$ we have $\alpha\beta=1$.
Additionally $\alpha^3-\beta^3=36$. By letting $\gamma=\alpha-\beta$ we have
$$ 36 = \gamma\left(\gamma^2+3\alpha\beta\right)=\gamma^3+3\gamma $$ hence $\gamma\in\mathbb{R}$ is a root of $x^3+3x-36=(x-3)\underbrace{(x^2+3x+12)}_{\text{negative discriminant}}$. This proves $\gamma=3$.