Problem 57 - Elementary and Intermediate Algebra

Question

Factor difference of cubes. \(64 t^{6}-27 v^{3}\)

Step-by-Step Solution

Answer

The factored form of the expression \(64 t^{6} - 27 v^{3}\) is \((4t^2 - 3v)(16t^4 + 12t^2v + 9v^2)\).

Step 1: Identify the Difference of Cubes Formula

The formula for factoring the difference of cubes is: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]We need to express the given expression \(64t^6 - 27v^3\) in the form of \(a^3 - b^3\).

Step 2: Rewrite Each Term as a Cube

Recognize the terms as cubes: \(64t^6 = (4t^2)^3\) and \(27v^3 = (3v)^3\). The expression can be rewritten as: \[ (4t^2)^3 - (3v)^3 \]

Step 3: Substitute into the Difference of Cubes Formula

Now substitute \(a = 4t^2\) and \(b = 3v\) into the difference of cubes formula: \[ (4t^2 - 3v)((4t^2)^2 + (4t^2)(3v) + (3v)^2) \]

Step 4: Simplify the Expression

Calculate each term in the expression:- \((4t^2)^2 = 16t^4\)- \((4t^2)(3v) = 12t^2v\)- \((3v)^2 = 9v^2\)The expression becomes:\[ (4t^2 - 3v)(16t^4 + 12t^2v + 9v^2) \]

Key Concepts

Polynomial Factorization

When dealing with algebraic expressions, polynomial factorization is a powerful tool to simplify and manipulate these expressions. Factorization involves rewriting a polynomial as a product of simpler polynomials. In simpler terms, it means breaking down a complex expression into simpler parts that multiply together to give the original expression.

• Factorization helps in solving polynomial equations by simplifying them.
• It reveals the roots (or zeros) of the polynomial, which are the values for which the polynomial equals zero.
• By understanding factorization, one can also sketch the graph of a polynomial function more easily.

Learning to factor different types of polynomials, such as quadratics, difference of squares, and difference of cubes, expands your toolkit for tackling complex algebraic expressions. In our example, we’re applying the special technique of 'difference of cubes' to factor the given polynomial. Identifying patterns in the terms of the polynomial is often the first step in successful factorization.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operations such as addition, subtraction, multiplication, and division. These expressions are foundational in algebra and are used to represent real-world situations in a concise mathematical form.

• Algebraic expressions can be simplified, evaluated, or factored.
• Expressions are denoted by variables such as 't' or 'v' in our exercise, which represent values that can change.
• Simplifying these expressions involves combining like terms and using arithmetic operations.

In tackling our difference of cubes problem, we identify components of the expression that can be rewritten and simplified. Understanding how to manipulate variables and exponents in algebraic expressions is key to solving these types of problems. By factorizing, we break down the expression into parts that are easier to work with, ultimately revealing new insights and solutions.

Difference of Cubes Formula

The difference of cubes formula is a specific polynomial identity used in factorization. It is particularly handy when you observe terms in an expression that are cubes of other terms. The formula itself is:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]

• This formula helps to quickly factor expressions comprised of cubes.
• Identifying 'a' and 'b' from the expression is the crucial first step.
• After identification, the expression can be systematically broken down using the formula.

In the exercise example, the expression \(64t^6 - 27v^3\) was rewritten in the form of \((4t^2)^3 - (3v)^3\). Recognizing these cube structures allows us to apply the formula smoothly. Implementing this, each component of \((4t^2 - 3v)(16t^4 + 12t^2v + 9v^2)\) can be derived, which helps simplify even more complex expressions efficiently. Knowing and applying this formula can be a significant advantage in algebra.