Problem 27 - Precalculus: Functions and Graphs

Question

Factor the polynomial. $$a^{6}-b^{6}$$

Step-by-Step Solution

Answer

The polynomial factors to \((a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2)\).

Step 1: Recognize the Difference of Squares

The given polynomial is \( a^6 - b^6 \), which is in the form \( x^2 - y^2 \). This can be factored using the difference of squares formula: \( x^2 - y^2 = (x - y)(x + y) \).

Step 2: Apply the Difference of Squares

Apply the difference of squares: \( a^6 - b^6 = (a^3)^2 - (b^3)^2 = (a^3 - b^3)(a^3 + b^3) \).

Step 3: Recognize the Sum and Difference of Cubes

The expression consists of two terms: \( a^3 - b^3 \) and \( a^3 + b^3 \). These can be factored further using the sum and difference of cubes formulas.

Step 4: Apply the Difference of Cubes Formula

The formula for the difference of cubes is \( x^3 - y^3 = (x-y)(x^2+xy+y^2) \). Therefore, \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).

Step 5: Apply the Sum of Cubes Formula

The formula for the sum of cubes is \( x^3 + y^3 = (x+y)(x^2-xy+y^2) \). Therefore, \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \).

Step 6: Write the Complete Factored Form

Combine the factors obtained to write the complete factored form of the original polynomial: \( a^6 - b^6 = (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2) \).

Key Concepts

Difference of Squares

The **difference of squares** is a fundamental concept in algebra that involves factoring expressions where two squares are subtracted. Let's start with the general formula, which is \( x^2 - y^2 = (x - y)(x + y) \). This can be used when you have two perfect squares separated by a minus sign. The given polynomial \( a^6 - b^6 \) fits this pattern. We can rewrite it as \((a^3)^2 - (b^3)^2\), recognizing that both \(a^3\) and \(b^3\) are perfect squares.

• The first step is to make each term a perfect square, which allows us to apply the difference of squares formula.
• Then, we factor it into two binomials: \((a^3 - b^3)(a^3 + b^3)\).

This creates an opportunity to further factor each resulting binomial using the concepts of difference and sum of cubes.

Difference of Cubes

Once we've factored our polynomial using the difference of squares, we are left with two terms: \(a^3 - b^3\) and \(a^3 + b^3\). The **difference of cubes** is a specific way to factor a polynomial of the form \( x^3 - y^3 \), and the formula to use is \( x^3 - y^3 = (x-y)(x^2+xy+y^2) \).

• In our exercise, applying this to \( a^3 - b^3 \), we get \( (a - b)(a^2 + ab + b^2) \).
• This formula helps break down more complex polynomials and is useful when dealing with cubic terms.

Understanding this concept allows us to simplify the polynomial to a more manageable expression, ensuring every factor is in its most simplified form.

Sum of Cubes

The **sum of cubes** formula is employed when we have an expression in the form of \( x^3 + y^3 \). It might seem slightly more difficult, but it's straightforward with the right formula: \( x^3 + y^3 = (x+y)(x^2-xy+y^2) \). This allows us to factor sums of cubes.

• Using this formula for \( a^3 + b^3 \) gives us \( (a + b)(a^2 - ab + b^2) \).
• Just like the difference of cubes, the sum of cubes simplifies expressions that at first seem not easily factored.

Combining these techniques, we arrive at a fully factored expression for the original polynomial \( a^6 - b^6 \): \( (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2) \). Understanding these processes makes complex polynomials much easier to handle.