Question

factor the following cubic.
$x^3 + 8$
a $(x + 2)(x^2 - 2x - 4)$
b $(x + 2)(x^2 - 2x + 4)$
c $(x + 2)(x^2 + 2x - 4)$
d $(x + 2)(x^2 + 2x + 4)$

Explanation:

Step1: Recall the sum of cubes formula

The sum of cubes formula is \(a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})\).

Step2: Identify \(a\) and \(b\) in \(x^{3}+8\)

For the expression \(x^{3}+8\), we can rewrite \(8\) as \(2^{3}\). So, \(a=x\) and \(b = 2\) since \(x^{3}+2^{3}\) is in the form of \(a^{3}+b^{3}\).

Step3: Apply the sum of cubes formula

Substitute \(a=x\) and \(b = 2\) into the formula \((a + b)(a^{2}-ab + b^{2})\). We get \((x + 2)(x^{2}-x\times2+2^{2})=(x + 2)(x^{2}-2x + 4)\).

Answer:

B. \((x + 2)(x^{2}-2x + 4)\)