factor: 8 + x³\na (2 + x)(4 - 2x + x²)\nb (2 + x)³\nc (2 - x)(4 + 2x + x²)\nd (2 + x)(4 + x²)\na² + 2ab +…

factor: 8 + x³\na (2 + x)(4 - 2x + x²)\nb (2 + x)³\nc (2 - x)(4 + 2x + x²)\nd (2 + x)(4 + x²)\na² + 2ab + b²=(a + b)²\na² - 2ab + b²=(a - b)²\na³ + b³=(a + b)(a² - ab + b²)\na³ - b³=(a - b)(a² + ab + b²)

factor: 8 + x³\na (2 + x)(4 - 2x + x²)\nb (2 + x)³\nc (2 - x)(4 + 2x + x²)\nd (2 + x)(4 + x²)\na² + 2ab + b²=(a + b)²\na² - 2ab + b²=(a - b)²\na³ + b³=(a + b)(a² - ab + b²)\na³ - b³=(a - b)(a² + ab + b²)

Answer

Answer:

A. $(2 + x)(4-2x + x^{2})$

Explanation:

Step1: Recall sum - of - cubes formula

$a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$

Step2: Identify values of a and b

For $8+x^{3}$, we have $a = 2$ (since $2^{3}=8$) and $b=x$.

Step3: Substitute values into formula

$(2)^{3}+x^{3}=(2 + x)(2^{2}-2x+x^{2})=(2 + x)(4-2x + x^{2})$