Problem 62 - Elementary and Intermediate Algebra

Question

Factor. \((b-c)^{3}-1,000\)

Step-by-Step Solution

Answer

The expression can be factored to \((b-c-10)(b^2 + 10b - 2bc - 10c + c^2 + 100)\).

Step 1: Recognize the Form

The given expression \((b-c)^{3}-1,000\) is a difference of cubes. Recall that a difference of cubes can be factored using the formula: \[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \] where in this exercise, \(a = (b-c)\) and \(b = 10\) since \(10^3 = 1000\).

Step 2: Factor Using Difference of Cubes Formula

Let \(a = (b-c)\) and \(b = 10\). Substitute these into the difference of cubes formula: \[ (b-c)^{3} - 10^{3} = ((b-c) - 10)((b-c)^{2} + (b-c) \cdot 10 + 10^{2}) \] Simplify this into two factors.

Step 3: Simplify the Binomial Factor

The first factor \(((b-c) - 10)\) simplifies to \(b-c-10\).

Step 4: Expand the Trinomial Factor

For the second factor, compute each part: - \((b-c)^{2} = b^2 - 2bc + c^2\)- \((b-c) \cdot 10 = 10b - 10c\)- Finally, \(10^{2} = 100\).Combine these to form: \[ b^2 - 2bc + c^2 + 10b - 10c + 100 \]

Step 5: Final Factored Form

Combine the simplified terms together:\[ (b-c-10)(b^2 + 10b - 2bc - 10c + c^2 + 100) \] is the fully factored form of the original expression.

Key Concepts

Difference of Cubes

When you encounter expressions such as \((b-c)^{3}-1000\), you're looking at a special case called "difference of cubes." This means both terms in the expression are perfect cubes. The general formula for factoring a difference of cubes is:\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]In this scenario, you can identify \(a = (b-c)\) and \(b = 10\). Remember, \(10^3 = 1000\), confirming that 1000 is indeed a cube.

• Identify a and b: Look at each component in the expression and determine what the cube roots are.
• Apply the formula: Simply substitute what you identified as \(a\) and \(b\) in the difference of cubes formula.

This will give you a form that's much easier to simplify further. This technique is a powerful tool for breaking down complex polynomials into manageable factors.

Binomial Factor

After recognizing the difference of cubes, you're left with a part of the problem that involves a "binomial factor." This is the first factor inside the parentheses: \((a-b)\).In our exercise, you've set \(a = (b-c)\) and \(b = 10\), leading to the binomial: \((b-c)-10\).

• Simplify the Expression: This translates into a straightforward subtraction to obtain \(b-c-10\).
• Why it's called a "binomial": "Bi-" in binomial refers to "two" components. Here, \(b-c\) and \(-10\) form the two components of this expression.

The binomial factor plays a critical role because it starts breaking down the original expression into simpler parts. Once it's simplified to \(b-c-10\), you can move on to other components with more ease.

Trinomial Expansion

After simplifying the binomial factor, the next task is to dive deeper into the "trinomial expansion." In this expression, it's the second factor from the original difference of cubes formula:\[ (b-c)^2 + (b-c) \cdot 10 + 100 \]Breaking it down further brings you to:

• \((b-c)^2\)expands to \(b^2 - 2bc + c^2\).
• \((b-c) \cdot 10\)becomes \(10b - 10c\).
• \(10^2\) simplifies to \(100\).

Combining these creates a trinomial:\[ b^2 - 2bc + c^2 + 10b - 10c + 100 \].This trinomial consists of three terms that include squares and products of diverse variables.Understanding this helps you see how multiple terms come together in the factorization, offering clarity and control over the polynomial structure.