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…

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

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

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 sum - of - cubes formula

Substitute $a = 2$ and $b=x$ into the formula $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. We get $(2 + x)(2^{2}-2x+x^{2})=(2 + x)(4-2x+x^{2})$.

Answer:

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