To identify which long division problem can be used to prove the formula for factoring the difference of two perfect cubes, we need to recall the formula for factoring such expressions:
The formula for the difference of two cubes is given by:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
This tells us that the difference of the cubes of two numbers [tex]\(a\)[/tex] and [tex]\(b\)[/tex] can be factored into the product of the difference of the two numbers and a quadratic polynomial involving the two numbers.
Given this, let's analyze the choices provided:
1. [tex]\(a - b \longdiv { a ^ { 2 } + a b + b ^ { 2 } }\)[/tex]
- Here, we are performing division where the divisor is [tex]\(a^2 + ab + b^2\)[/tex]. This polynomial is part of the result we get after factoring, not the entire expression we need to consider for the proof of the factorization.
2. [tex]\(a + b \longdiv { a ^ { 2 } - a b + b ^ { 2 } }\)[/tex]
- Similar to the first option, this division involves a polynomial similar to part of the factorization of a sum of cubes, specifically [tex]\(a^2 - ab + b^2\)[/tex], which isn't applicable for the difference of cubes.
3. [tex]\(a + b \sqrt{a^3+0 a^2 b+0 a b^2-b^3}\)[/tex]
- This notation mixes a radical expression with the sum and difference of terms in a way that isn’t applicable to our straightforward polynomial factorization method.
4. [tex]\(a - b \longdiv { a ^ { 3 } + 0 a ^ { 2 } b + 0 a b ^ { 2 } - b ^ { 3 } }\)[/tex]
- Here, we see that the division corresponds to the expression [tex]\(a^3 - b^3\)[/tex], with [tex]\(a - b\)[/tex] as the divisor. This perfectly aligns with our factorization formula for the difference of cubes:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
- When [tex]\(a^3 - b^3\)[/tex] is divided by [tex]\(a - b\)[/tex], the quotient is indeed [tex]\(a^2 + ab + b^2\)[/tex], thereby verifying the factorization formula.
Therefore, this is the correct option that can be used to prove the formula for factoring the difference of two perfect cubes:
[tex]\[ \boxed{a - b \longdiv { a ^ { 3 } + 0 a ^ { 2 } b + 0 a b ^ { 2 } - b ^ { 3 } }} \][/tex]
