Question

[SOLVED] Let \(\alpha := \sqrt{2-\sqrt[3]{5}} \in \mathbb{R}\) Show that \(\alpha\) is algebraic over \(\mathbb{Q}\) by finding a polynomial \(p(x)\in \mathbb{Q}[x]\) such that \(p(\alpha)=0\)

Is there a better way to do this problem other than to find \(\alpha^6\) and then use guess and check with lower powers until you find the correct coefficients?

Answer 1

\[p(x)=x^6-6x^4+12x^2-3\]

Answer 2

What would be the most effective way of finding that polynomial?

Answer 3

let \[x=\sqrt{2-\sqrt[3]{5}}\]

\[x^2=2-\sqrt[3]{5}\]

\[x^2-2=-\sqrt[3]{5}\]

\[(x^2-2)^3=-5\]
\[(x^2-2)^3+5=0\]

\[(x^2-2)^3+5=x^6-6x^4+12x^2-3\]

Answer 4

Awesome! Thanks.

Answer 5

yw

Answer 6

So if it were a cubed root on the outside, instead of squaring first, I would cube \(\alpha\), get all the rational numbers and \(\alpha^3\) on one side, and then cube again to the \(-\sqrt[3]{5}\) into a rational number. And when that's all done, solve for 0.

Answer 7

yes

Answer 8

Thank you.