# 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.

$\blacksquare$