factor: 8 + x^3\na (2 + x)(4 - 2x + x^2)\nb (2 + x)^3\nc (2 - x)(4 + 2x + x^2)\nd (2 + x)(4 + x^2)\na^2 +…

factor: 8 + x^3\na (2 + x)(4 - 2x + x^2)\nb (2 + x)^3\nc (2 - x)(4 + 2x + x^2)\nd (2 + x)(4 + x^2)\na^2 + 2ab + b^2=(a + b)^2\na^2 - 2ab + b^2=(a - b)^2\na^3 + b^3=(a + b)(a^2 - ab + b^2)\na^3 - b^3=(a - b)(a^2 + ab + b^2)

factor: 8 + x^3\na (2 + x)(4 - 2x + x^2)\nb (2 + x)^3\nc (2 - x)(4 + 2x + x^2)\nd (2 + x)(4 + x^2)\na^2 + 2ab + b^2=(a + b)^2\na^2 - 2ab + b^2=(a - b)^2\na^3 + b^3=(a + b)(a^2 - ab + b^2)\na^3 - b^3=(a - b)(a^2 + ab + b^2)

Answer

Explanation:

Step1: Identify the sum - of - cubes formula

We know that $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. Here, $8+x^{3}$ can be written as $2^{3}+x^{3}$, where $a = 2$ and $b=x$.

Step2: Apply the formula

Substitute $a = 2$ and $b=x$ into the sum - of - cubes formula. We get $(2 + x)(2^{2}-2x+x^{2})=(2 + x)(4-2x + x^{2})$.

Answer:

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