factorize x^3-8

1) Let's factor in

ℝ

x^{3} - 8

⇥1.1) The roots of

x^{3} - 8

are the solutions of

x^{3} = 8

⇥The complex solutions are:

\sqrt[3]{8} \cdot \exp (i \cdot \frac{2 k π}{3})

, with k ranging from 0 to 2

⇥

k	Exponential solution	Transformed and simplified as
0	$2 \exp (i \cdot \frac{2 \cdot 0 \cdot π}{3}) = 2$	$2 \cdot (\cos (0) + i \cdot \sin (0)) = 2$
1	$2 \exp (i \cdot \frac{2 \cdot 1 \cdot π}{3}) = 2 \exp (\frac{i \cdot 2 π}{3})$	$2 \cdot (\cos (\frac{2 π}{3}) + i \cdot \sin (\frac{2 π}{3})) = - 1 + i \cdot \sqrt{3}$
2	$2 \exp (i \cdot \frac{2 \cdot 2 \cdot π}{3}) = 2 \exp (\frac{i \cdot 4 π}{3})$	$2 \cdot (\cos (\frac{4 π}{3}) + i \cdot \sin (\frac{4 π}{3})) = - 1 - i \cdot \sqrt{3}$

⇥The factorization for the real solution(s) yields:

x - 2

⇥1.2) Perform the Euclidean division of

(x^{3} - 8)

(x - 2)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{3} - 8$	$\frac{x^{3}}{x}$	$x^{2}$	$x^{2} \cdot (x - 2) = x^{3} - 2 x^{2}$	$(x^{3} - 8) - (x^{3} - 2 x^{2}) = 2 x^{2} - 8$
$2 x^{2} - 8$	$\frac{2 x^{2}}{x}$	$2 x$	$2 x \cdot (x - 2) = 2 x^{2} - 4 x$	$(2 x^{2} - 8) - (2 x^{2} - 4 x) = 4 x - 8$
$4 x - 8$	$\frac{4 x}{x}$	$4$	$4 \cdot (x - 2) = 4 x - 8$	$(4 x - 8) - (4 x - 8) = 0$

⇥So the result of the Euclidean division is:

x^{2} + 2 x + 4

ⓘ The other roots,

- 1 - i \cdot \sqrt{3}

- 1 + i \cdot \sqrt{3}

, are not real:

x^{2} + 2 x + 4

cannot be factored in ℝ

▶Factors found:

(x - 2) \cdot (x^{2} + 2 x + 4)

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