Prove that $\sqrt[3]{45+29\sqrt{2}} +\sqrt[3]{45-29\sqrt{2}} $ is rational.

Question

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!

Answer 1

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 $x=6$.

Answer 2

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.

Answer 3

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.

Answer 4

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, $m=3$

$$\implies29=3(3b)^2+2b^3\implies b^3=1\implies b=1, a=?$$