Show that $x$ is an algebraic number? Where $x$ is...

Question

Can someone help me with the following problem? Show that $x=\sqrt2+\sqrt[3]3$ is an algebraic number. By finding a polynomial with rational coefficients for which $x$ is a root of.

Can someone please find this polynomial and give me some advice on how to come up with such a polynomial. Thanks

Answer 1

Start with $x - \sqrt 2 = \sqrt[3] 3$. Cube both sides and expand to get $- \sqrt 2 (3x^2 + 2) + x^3 + 6x = 3$. Again, write this as $- \sqrt 2 (3x^2 + 2) = - x^3 - 6x + 3$ and square. This will give you a polynomial with $x$ as a root.

As an additional exercise, show that the polynomial is irreducible to verify that the degree of $x$ is $6$.

Answer 2

Isolate $\sqrt[3]3$ on one side. Then raise to the third power. Then isolate $\sqrt 2$ on one side. Then square both sides.