Problem 51 - College Algebra

Question

In Exercises 49-52, use the Binomial Theorem to expand each expression and write the result in simplified form. $$\left(x^{\frac{1}{3}}-x^{-\frac{1}{3}}\right)^{3}$$

Step-by-Step Solution

Answer

The simplified form of the expression \( \left(x^{\frac{1}{3}} - x^{-\frac{1}{3}}\right)^3 \) is \( x-3x^{\frac{1}{3}}+3x^{-\frac{1}{3}}-x^{-1} \)

Step 1: Identify the components of the binomial expression

In our case the binomial expression is \( \left(x^{\frac{1}{3}} - x^{-\frac{1}{3}}\right)^3 \), so, \( a = x^{\frac{1}{3}} \), \( b = -x^{-\frac{1}{3}} \), and \( n = 3 \)

Step 2: Apply the Binomial Theorem

The Binomial Theorem is expressed as \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \). Applying this to our expression we get: \( x+3x^{\frac{2}{3}}(-x^{-\frac{1}{3}})+3x^{\frac{1}{3}}(-x^{-\frac{1}{3}})^2+(-x^{-\frac{1}{3}})^3 \)

Step 3: Simplify the expression

Simplifying, we get: \( x-3x^{\frac{1}{3}}+3x^{-\frac{1}{3}}-x^{-1} \)

Key Concepts

Binomial Expansion

The binomial expansion is a powerful tool in algebra that allows us to simplify expressions raised to a power. You start with a binomial, which is simply an algebraic expression with two terms, like \((a + b)^n\). With the Binomial Theorem, we can expand this binomial into a sum of terms, which is much simpler to work with.

The key formula in the binomial theorem is: \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\] Here, \(\binom{n}{k}\) denotes the binomial coefficient, representing how many ways you can choose \(k\) objects out of \(n\). Each term in the expansion involves the power \(n-k\) of \(a\) and the power \(k\) of \(b\), multiplied by this binomial coefficient.

In our specific exercise, the binomial \((x^{\frac{1}{3}} - x^{-\frac{1}{3}})^3\) can be expanded using the same principle. Each step uses the coefficients and powers of the terms to build the full expansion.

Fractional Exponents

Fractional exponents are a convenient way to express roots in algebra. Instead of writing \(\sqrt{x}\), you can write \(x^{\frac{1}{2}}\), which is clearer to work with in many algebraic operations.

The expression \(x^{\frac{1}{3}}\) tells us to take the cube root of \(x\). Similarly, \(x^{-\frac{1}{3}}\) stands for the reciprocal of the cube root of \(x\). When you deal with fractional exponents:

• \(x^{\frac{m}{n}} = (x^m)^{\frac{1}{n}}\) or \((\sqrt[n]{x})^m\)
• If \(m\) is 1, then it is just the nth root: \(x^{\frac{1}{n}} = \sqrt[n]{x}\)
• If the exponent is negative, like \(-\frac{1}{n}\), it means \(\frac{1}{\sqrt[n]{x}}\)

Understanding fractional exponents is essential when expanding expressions and performing algebraic simplifications, as seen in \((x^{\frac{1}{3}} - x^{-\frac{1}{3}})^3\). It tells you how to manipulate the roots during calculation.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations like addition, subtraction, multiplication, and division. They serve as the building blocks of algebra, allowing you to encode numeric relationships into symbolic form.

Let's break down the expression from our exercise: \((x^{\frac{1}{3}} - x^{-\frac{1}{3}})^3\). This expression is formed by:

• The base terms \(x^{\frac{1}{3}}\) and \(-x^{-\frac{1}{3}}\), both containing variables and powers.
• An operation (subtraction) between these two terms.
• A power of 3, indicating that we multiply the entire expression by itself three times.

In algebra, you perform operations on expressions like these to simplify or evaluate them, often using rules from arithmetic and algebra. Recognizing the structure within algebraic expressions and knowing how to apply algebraic principles, such as the Binomial Theorem, helps in effectively tackling exercises in algebra.