To factor the expression [tex]\(64 g^3 + 8\)[/tex], we start by looking for a common factor in each term. Notice that both terms, [tex]\(64 g^3\)[/tex] and [tex]\(8\)[/tex], can be divided by 8. Therefore, we factor out 8 first:
[tex]\[ 64 g^3 + 8 = 8 (8 g^3 + 1) \][/tex]
Next, we recognize that [tex]\(8 g^3 + 1\)[/tex] can be treated as a sum of cubes. Recall the sum of cubes formula:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
In our case, [tex]\(8 g^3 = (2g)^3\)[/tex] and [tex]\(1 = 1^3\)[/tex]. So we can rewrite the expression as:
[tex]\[ 8 g^3 + 1 = (2g)^3 + 1^3 \][/tex]
Applying the sum of cubes formula, we get:
[tex]\[ (2g)^3 + 1^3 = (2g + 1) \left( (2g)^2 - (2g)(1) + 1^2 \right) \][/tex]
Now we simplify the components inside the parentheses:
[tex]\[ (2g)^2 = 4g^2 \][/tex]
[tex]\[ (2g)(1) = 2g \][/tex]
[tex]\[ 1^2 = 1 \][/tex]
Putting it all together, the sum of cubes factors into:
[tex]\[ (2g + 1)(4g^2 - 2g + 1) \][/tex]
So the original expression can be factored as:
[tex]\[ 8 (8 g^3 + 1) = 8 (2g + 1)(4g^2 - 2g + 1) \][/tex]
This means the factored form of [tex]\(64 g^3 + 8\)[/tex] is:
[tex]\[ 8 (2g + 1)(4g^2 - 2g + 1) \][/tex]
Reviewing the given options:
1. [tex]\((4 g+2)\left(16 g^2+8 g-4\right)\)[/tex]
2. [tex]\((4 g+2)\left(16 g^2-8 g-4\right)\)[/tex]
3. [tex]\((4 g+2)\left(16 g^2+8 g+4\right)\)[/tex]
4. [tex]\((4 g+2)\left(16 g^2-8 g+4\right)\)[/tex]
None of these options directly match our factored form. Notice, however, that each factor inside these options can be written differently, but result still matches algebraically. Let's align properly;
Comparing [tex]\((4 g+2)\)[/tex] with [tex]\(2(2g+1)\)[/tex] and [tex]\(8\)[/tex] factored from our complete solution.
So, we establish factored form derived accurately:
The closest (optionally confirming could be their distributed results). But exact factors based provided;
Best Final: \( ( 2g+ 1 )( 4g^2 - 2g+ 1 ) accurate aligned. ) without any error using standard/codes.
Thus---
Which informs correctly; right assembly approach / teaching required imparts accurate expressions followed method.
