Problem 111 - Intermediate Algebra

Question

Simplify each expression. $$ \frac{a^{6}-64}{\left(a^{2}+2 a+4\right)\left(a^{2}-2 a+4\right)} $$

Step-by-Step Solution

Answer

The expression simplifies to \( a^2 - 4 \).

Step 1: Recognize the Structure of Difference of Squares

To simplify the expression \( \frac{a^{6}-64}{(a^{2}+2a+4)(a^{2}-2a+4)} \), begin by identifying if the numerator \( a^6 - 64 \) can be expressed using the difference of squares formula. Rewrite \( 64 \) as \( 2^6 \), so the expression becomes \( a^6 - 2^6 \).

Step 2: Apply the Difference of Squares Formula

Recall the difference of squares formula: \( x^2 - y^2 = (x-y)(x+y) \). For \( a^6 - 2^6 \), think of it as \( (a^3)^2 - (2^3)^2 \). Apply the formula to factor as \( (a^3 - 2^3)(a^3 + 2^3) \).

Step 3: Simplify Further Using Sum and Difference of Cubes

The factored form \( (a^3 - 2^3)(a^3 + 2^3) \) can be further simplified using cube identities: for any \( x^3 - y^3 = (x-y)(x^2 + xy + y^2) \) and \( x^3 + y^3 = (x+y)(x^2 - xy + y^2) \). So it becomes: \((a-2)(a^2 + 2a + 4)(a+2)(a^2 - 2a + 4)\).

Step 4: Cancel Common Factors

Now the complete factored form of the numerator \( (a-2)(a^2 + 2a + 4)(a+2)(a^2 - 2a + 4) \) includes terms present in the denominator \( \left(a^{2}+2a+4\right)(a^{2}-2a+4) \). Therefore, cancel the common terms: \( a^2+2a+4 \) and \( a^2-2a+4 \).

Step 5: Conclusion to Simplified Expression

After cancellation, simplifying the entire expression results in: \((a-2)(a+2)\). The simplified expression is \( a^2 - 4 \).

Key Concepts

Difference of Squares

The concept of difference of squares is a very handy tool when simplifying algebraic expressions. This method involves identifying patterns that fit into the formula: \[ x^2 - y^2 = (x-y)(x+y) \] This formula works whenever you have two terms, each a perfect square, separated by a subtraction sign.
In our original problem, we start with the numerator \( a^6 - 64 \), which does not initially appear as a difference of squares. By recognizing that both \( a^6 \) and \( 64 \) can be expressed as squares—\((a^3)^2\) and \((2^3)^2\)—we can then apply the difference of squares factoring.
Transforming \( a^6 - 64 \) into \((a^3)^2 - (2^3)^2\) reveals the structure, allowing us to factor it into: - \((a^3 - 2^3)(a^3 + 2^3)\) This technique is a useful stepping stone to further simplification using other algebraic identities.

Sum and Difference of Cubes

Algebra also gives us tools like the sum and difference of cubes identities, which help us factor expressions of the form:

• \( x^3 + y^3 = (x+y)(x^2 - xy + y^2) \)
• \( x^3 - y^3 = (x-y)(x^2 + xy + y^2) \)

In our expression, having already transformed \( a^6 - 64 \) into \((a^3 - 2^3)(a^3 + 2^3)\), we can now dive deeper using these identities.
Applying the difference of cubes formula to \( a^3 - 2^3 \) gives us:

• \( (a - 2)(a^2 + 2a + 4) \)

Similarly, applying the sum of cubes formula to \( a^3 + 2^3 \) transforms it into:

• \( (a + 2)(a^2 - 2a + 4) \)

This level of factorization broadens our capability to simplify complex expressions by breaking them down into more manageable pieces.

Factoring Polynomials

Factoring polynomials is all about breaking down an expression into a product of simpler expressions or "factors." Through factoring, we can express a complicated polynomial as a multiplication of polynomials of lower degrees.
In the given problem, the entire process starts by identifying potential patterns for factoring in both the numerator and denominator. After applying the difference of squares method to the numerator and acquiring its factored form through the sum and difference of cubes: - \((a-2)(a^2 + 2a + 4)(a+2)(a^2 - 2a + 4)\) Factoring allows us to reduce the expression further by canceling identical terms present both in the numerator and the denominator:

• \(a^2 + 2a + 4\) cancels with itself.
• \(a^2 - 2a + 4\) cancels with itself.

We are left with: - \((a-2)(a+2)\)
These steps result in the simpler form of the expression, which highlights the use of factoring as a critical tool in algebra.