algebra precalculus - Prove $sqrt{2} + sqrt[3]{2}$ is irrational

I solved the problem by way of contradiction. Suppose $x = \sqrt{2} + \sqrt[3]{2}$ is rational. Then we have

$$2 = (x - \sqrt{2})^3 = x^3 - 3\sqrt{2}x^2 + 6x - 2\sqrt{2} = (x^3 + 6x) - \sqrt{2}(3x^2 - 2)$$
I've already shown that $\sqrt{2}$ is irrational, and it's easy to show a rational number plus an irrational number is irrational and that the product of an irrational and a rational is irrational. Given those facts, we have $(x^3 + 6x) \in \mathbb{Q}$ and $\sqrt{2}(3x^2 - 2) \not\in \mathbb{Q}$, so
$$2 = a\in\mathbb{Q} + b\not\in\mathbb{Q} \implies 2 \not\in\mathbb{Q}$$ a contradiction.

However, the book does it differently: it says

Show that $x$ satisfies an equation of the type $$x^6 + a_1x^5 + \ldots + a_6 = 0$$ where $a_1,\ldots,a_6$ are integers; prove that $x$
is then either irrational or an integer.

1. Is my way correct as well?


2. I don't understand how the book does it. How does one obtain that equation, and once obtained, how does it prove $x$ is irrational?

Answer

You've done half the job for their way: from
$$2 = (x - \sqrt{2})^3 = x^3 - 3\sqrt{2}x^2 + 6x - 2\sqrt{2} = (x^3 + 6x) - \sqrt{2}(3x^2 - 2)$$
you deduce: $$(x^3 + 6x-2)^2=2(3x^2+2)^2,$$

whence $$x^6-6x^4-4x^3+12x^2-24x-4=0.$$
By the Rational roots theorem, we know that a rational root has to be an integer, and a divisor of $4$, i. e. it can be only $\pm 1, \pm 2, \pm 4$. Just test them to check none is a root.