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:

\(\ds x - \sqrt 3\)

\(=\)

\(\ds \sqrt [3] 2\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x - \sqrt 3}^3\)

\(=\)

\(\ds 2\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x - \sqrt 3} \paren {x - \sqrt 3}^2\)

\(=\)

\(\ds 2\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x - \sqrt 3} \paren {x^2 - 2 x \sqrt 3 + 3}\)

\(=\)

\(\ds 2\)

\(\ds \leadsto \ \ \)

\(\ds x^3 - 3 \sqrt 3 x^2 + 9 x - 3 \sqrt 3\)

\(=\)

\(\ds 2\)

\(\ds \leadsto \ \ \)

\(\ds x^3 + 9 x - 2\)

\(=\)

\(\ds 3 \sqrt 3 x^2 + 3 \sqrt 3\)

\(\ds \)

\(=\)

\(\ds 3 \sqrt 3 \paren {x^2 + 1}\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x^3 + 9 x - 2}^2\)

\(=\)

\(\ds 3^2 {\sqrt 3}^2 \paren {x^2 + 1}^2\)

\(\ds \leadsto \ \ \)

\(\ds x^6 + 18 x^4 - 4 x^3 + 81 x^2 - 36 x + 4\)

\(=\)

\(\ds 27 \paren {x^4 + 2 x^2 + 1}\)

\(\ds \)

\(=\)

\(\ds 27 x^4 + 54 x^2 + 27\)

\(\ds \leadsto \ \ \)

\(\ds x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23\)

\(=\)

\(\ds 0\)

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.

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $158 \ \text{(a)}$