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Let $x = \sqrt[3]{3} + \sqrt{5}$
Notice that $(x+y)^2 = x^2 + 2xy + y^2$
Then, $x^2 = (\sqrt[3]{3} + \sqrt{5})^2 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5} + 5$
Then, $x^3 - 5 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5}$

Or notice that $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$
Then, $x^3 = (\sqrt[3]{3} + \sqrt{5})^3 = 3 + 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Then, $x^3 - 3 = 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$

Which I think the degree of the minial polynomial of $x$ over $\mathbb{Q}$ should have a degree of $6$.
Which means the minimal polynomial can be written as $x^6 + a_5 x^5 + \cdots + a_0$ for some $a_5, ..., a_0 \in \mathbb{Q}$.
But I just don't how get it from both the ways I provide above.
Or that $\sqrt[3]{3} + \sqrt{5}$ is not algebraic. So, I can not write such minimal polynomial out algebraic ?

José Carlos Santos
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2 Answers2

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If $x=\sqrt[3]3+\sqrt5$, then $\left(x-\sqrt5\right)^3=3$. In other words $x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0$. But\begin{align}x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0&\iff x^3+15x-3=(3x^2-5)\sqrt5\\&\implies(x^3+15x-3)^2=5(3x^2-5)^2\\&\iff x^6-15 x^4-6 x^3+375 x^2-90 x-116=0.\end{align}

José Carlos Santos
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  • We also need to show that the polynomial is irreducible. But I think we can directly prove that $[\mathbb {Q} (\sqrt[3]{3}+\sqrt{5}):\mathbb {Q}] =6$. – Paramanand Singh May 20 '21 at 02:28
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Set of algebraic numbers form a field, so the sum of two algebraic numbers $\sqrt[3]{3} + \sqrt{5}$ is again an algebraic number.

To find the minimal polynomial, you want to continue in either of the two ways that you have started.

If it is the first way, you have to take the suitable powers of $x^2-5$ and continue until getting a rational polynomial.

sabeelmsk
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