which is the factorization of $x^{3}+8$?\n$(x + 2)(x^{2}-2x + 4)$\n$(x - 2)(x^{2}+2x + 4)$\n$(x +…
which is the factorization of $x^{3}+8$?\n$(x + 2)(x^{2}-2x + 4)$\n$(x - 2)(x^{2}+2x + 4)$\n$(x + 2)(x^{2}-2x + 8)$\n$(x - 2)(x^{2}+2x + 8)$
Answer
Answer:
A. $(x + 2)(x^{2}-2x + 4)$
Explanation:
Step1: Recall 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 $b$
For $x^{3}+8$, we have $a=x$ and $b = 2$ since $8=2^{3}$.
Step3: Apply the formula
Substitute $a=x$ and $b = 2$ into the sum - of - cubes formula: $x^{3}+8=x^{3}+2^{3}=(x + 2)(x^{2}-2x + 4)$.