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

## Example of Algebraic Number

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

## Proof

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

We have that:

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

Thus $\sqrt [3] 2 + \sqrt 3$ is a root of $x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23 = 0$.

Hence the result by definition of algebraic number.

$\blacksquare$