Prove that $\sqrt[3]{45+29\sqrt{2}} +\sqrt[3]{45-29\sqrt{2}} $ is rational.
So I assumed, $\sqrt[3]{45+29\sqrt{2}} +\sqrt[3]{45-29\sqrt{2}} = x$
So we have to prove $x$ is rational, I did so and cubed $x$, so I got:
$45+29\sqrt{2} +{45-29\sqrt{2}} + 21x = x^3$
And on further solving I got:
$$x^3-21x-90=0$$
What to do further? I'm unable to evaluate roots so it is probably wrong. Any better methods? Thanks!
What you did is fine. It turns out that $6$ is a root of that polynomial. You can guess it, using the rational root theorem. Besides,$$x^3-21x-90=(x-6)(x^2+6x+15).$$So, it has no other real roots, and therefore $x=6$.
You can apply this method,
Let
$$a, b\in\mathbb Z$$ then
$$\begin{align}&\begin{cases} (a+b\sqrt 2)^3=45+29\sqrt 2 \\ (a-b\sqrt 2)^3=45-29\sqrt 2 \end{cases} \\ \\ \iff &\begin{cases} a^3 +6ab^2+(3a^2b+2b^3)\sqrt 2=45+29\sqrt 2 \\a^3 +6ab^2-(3a^2b+2b^3) \sqrt 2=45-29\sqrt 2 \end{cases}\end{align} $$
You get,
$$\begin{cases} a^3+6ab^2=45 \\ 3a^2b+2b^3=29 \end{cases}$$
Don't solve the system of equations.
$$2b^3+3a^2b-29=0$$
$29$ is a prime number. Just apply Rational Root Theorem, you need $b=±1,b=±29$.
Thus, we have
$$a=3, ~b=1$$
Finally,
$$45±29\sqrt 2=(3±\sqrt 2)^3$$
The rest of solution is easy.
This is just a supplement to José Carlos Santos' nice answer which wants to draw your attention to the general solution of cubic equations.
You showed that $\xi = \sqrt[3]{45+29\sqrt{2}} +\sqrt[3]{45-29\sqrt{2}}$ is a solution of $x^3-21x-90=0$. José Carlos Santos gave a simple proof that we must have $\xi = 6$.
Anyway, you realized that there is a relationship between expressions like $\sqrt[3]{45+29\sqrt{2}} +\sqrt[3]{45-29\sqrt{2}}$ and solutions of cubic equations. Cubic equations can be solved by the Cardano formula. See here.
If you know the Cardano formula, then you immediately see that an expression like $\sqrt[3]{45+29\sqrt{2}} +\sqrt[3]{45-29\sqrt{2}}$ must be the single real solution of a cubic equation. You can determine this equation as you did. If it has a rational solution $x$, we know that $x$ must an integer dividing the constant term $-90$, and this can be checked by trying.
However, I admit that José Carlos Santos approach is much easier: He checks whether a rational solution exists and then uses polynomial division to show that the other two solutions are non-real.
Like @loneStudent,
put $a=mb$ (where $m$ is real) in $$\dfrac{a^3+6ab^2}{3a^2b+2b^3}=\dfrac{45}{29}$$
to find $$\dfrac{m^3+6m}{3m^2+2}=\dfrac{45}{29}\iff29m^3-135m^2+174m-90=0$$
$29m^3-135m^2+174m-90=29m^2(m-3)-48m(m-3)+30(m-3)=(m-3)(29m^2-48m+30)$
As the discriminant $48^2-4\cdot29\cdot30=24(96-5\cdot29)<0,$ and $m$ is real, $m=3$
$$\implies29=3(3b)^2+2b^3\implies b^3=1\implies b=1, a=?$$
