Algebraic Number/Examples/Cube Root of 2 plus Root 3

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Example of Algebraic Number

$\sqrt [3] 2 + \sqrt 3$ is an algebraic number.


Proof

Let $x = \sqrt [3] 2 + \sqrt 3$.

We have that:

\(\displaystyle x - \sqrt 3\) \(=\) \(\displaystyle \sqrt [3] 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x - \sqrt 3}^3\) \(=\) \(\displaystyle 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x - \sqrt 3} \paren {x- \sqrt 3}^2\) \(=\) \(\displaystyle 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x - \sqrt 3} \paren {x^2 - 2x\sqrt 3 + 3}\) \(=\) \(\displaystyle 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^3 - 3 \sqrt 3 x^2 + 9 x - 3 \sqrt 3\) \(=\) \(\displaystyle 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^3 + 9 x - 2\) \(=\) \(\displaystyle 3 \sqrt 3 x^2 + 3 \sqrt 3\)
\(\displaystyle \) \(=\) \(\displaystyle 3 \sqrt 3 \paren {x^2 + 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x^3 + 9 x - 2}^2\) \(=\) \(\displaystyle 3^2 {\sqrt 3}^2 \paren {x^2 + 1}^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^6 + 18 x^4 - 4 x^3 + 81 x^2 - 36 x + 4\) \(=\) \(\displaystyle 27 \paren {x^4 + 2 x^2 + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 27 x^4 + 54 x^2 + 27\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23\) \(=\) \(\displaystyle 0\)


Thus $\sqrt [3] 2 + \sqrt 3$ is a root of a non-zero polynomial with rational coefficients, namely $x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23 = 0$.

Hence the result by definition of algebraic number.

$\blacksquare$


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