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Determine if a real number is rational using roots of polynomial V


Our main tool is Corollary 2.3.

Solution: First, we need to find an equation with integer coefficients such that $(3+\sqrt{2})^{2/3}$ is a solution to it.

Let $x=(3+\sqrt{2})^{2/3}$. Taking cube for both sides, we obtain that $x^3=(3+\sqrt{2})^{2}$. Hence
$$
x^3=9+6\sqrt 2+2=11+6\sqrt 2.
$$ In particular, $x^3-11=6\sqrt 2$. Squaring both sides, we get
$$
(x^3-11)^2=72,
$$ which is
$$
x^6-22x^3+49=0.
$$ Consider the rational solution of $x^6-22x^3+49=0$. It follows from Corollary 2.3 that those rationa solutions can only be $\pm 1$,$\pm 7$, $\pm 49$.

Direct computations would show that none of them will be a solution of $x^6-22x^3+49=0$. Hence $x^6-22x^3+49=0$ has no rational solutions. Therefore, as a solution to $x^6-22x^3+49=0$, $(3+\sqrt{2})^{2/3}$ is irrational.


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