# Roots and multiplicity of the polynomial x^3 – 2

Calculate all roots and their multiplicity of $$X^3-2$$ in the rings $$\mathbb{Q}[X], \mathbb{R}[X], \mathbb{C}[X]$$.

## Solution

1)

$$!X^3 – 2 = 0 \quad \Leftrightarrow \quad X^3 = 2$$

has no solution in ℚ.

2)

$$!X^3 – 2 = 0 \quad \Rightarrow \quad \lambda_1 = \sqrt[3]{2}$$

is a solution.

There is a polynomial $$f \in \mathbb{R}[X]$$ with $$X^3-2 = (X-\sqrt[3]{2}) f$$.

$$!(X^3-2) / (X-\sqrt[3]{2}) = X^2+\sqrt[3]{2}X+\sqrt[3]{4} = f$$

f has degree 2, the roots can be found by the quadratic formula p-q-Formel:

$$!0 = X^2+\sqrt[3]{2}X+\sqrt[3]{4}\\\\ \Rightarrow \lambda_{2,3} = -\frac{\sqrt[3]{2}}{2} \pm \sqrt{\frac{\sqrt[3]{4}}{4}-\sqrt[3]{4}}$$

Under the radical sign is a negative number, so the $$\lambda_{2,3}$$ are no real numbers, therefore f is irreducible over ℝ.

We have: $$X^3-2$$ has one root, $$\lambda_1$$, with multiplicity 1.

3)

$$0 = X^2+\sqrt[3]{2}X+\sqrt[3]{4}$$ has two solutions in ℂ:

$$!\lambda_{2,3} = -\frac{\sqrt[3]{2}}{2} \pm \frac{\sqrt[6]{108}}{4}i$$

We have: $$X^3-2 = (X-\lambda_1)(X-\lambda_2)(X-\lambda_3)$$ has three roots with multiplicity 1.