Problem 16 - Contemporary Precalculus

Question

Compute and simplify. $$\left(x^{1 / 3}+y^{1 / 3}\right)\left(x^{2 / 3}-x^{1 / 3} y^{1 / 3}+y^{2 / 3}\right)$$

Step-by-Step Solution

Answer

Question: Multiply and simplify the expression: $(x^{1/3} + y^{1/3})(x^{2/3} - x^{1/3}y^{1/3} + y^{2/3})$ Answer: The simplified expression is: $x - x^{2/3}y^{1/3} + x^{1/3}y^{2/3}$

Step 1: Distribute the terms

We start by distributing each term of the first parenthetical with each term of the second parenthetical. $$\left(x^{1 / 3}\right)\left(x^{2 / 3}\right) - \left(x^{1 / 3}\right)\left(x^{1 / 3} y^{1 / 3}\right) + \left(x^{1 / 3}\right)\left(y^{2 / 3}\right) + \left(y^{1 / 3}\right)\left(x^{2 / 3}\right) - \left(y^{1 / 3}\right)\left(x^{1 / 3} y^{1 / 3}\right) + \left(y^{1 / 3}\right)\left(y^{2 / 3}\right)$$

Step 2: Simplify the terms

Now, we can simplify each term by adding the exponents of the same bases and apply the product rule for exponents, where the exponents of powers being multiplied are added. $$x^{1/3 + 2/3} - x^{1/3 + 1/3}y^{1/3} + x^{1/3}y^{2/3} + x^{2/3}y^{1/3} - x^{1/3}y^{1/3}y^{1/3} + y^{1/3+2/3}$$

Step 3: Perform additions in the exponents

Next, we perform the additions in the exponents and simplify the result. $$x^1 - x^{2/3}y^{1/3} + x^{1/3}y^{2/3} + x^{2/3}y^{1/3} - x^{1/3}y^{2/3} + y^1$$

Step 4: Combine like terms

Now, we can combine the like terms in the expression. $$x^1 - x^{2/3}y^{1/3} + x^{1/3}y^{2/3} + 0 - y^{1/3}y^{2/3} + y^1$$

Step 5: Simplify the sum

Once again, we use the product rule for exponents and simplify the expression. $$x^1 - x^{2/3}y^{1/3} + x^{1/3}y^{2/3} - y^1 + y^1 = x - x^{2/3}y^{1/3} + x^{1/3}y^{2/3}$$ So the simplified expression is: $$x - x^{2/3}y^{1/3} + x^{1/3}y^{2/3}$$

Key Concepts

Exponents and Radicals

Understanding exponents and radicals is crucial when working with algebraic expressions, as they frequently appear in mathematics. An exponent, such as in the term \(x^{m/n}\), indicates that the base \(x\) is to be raised to a power. In fractional exponents, the numerator represents the power to which the base is raised, and the denominator represents the root that is taken. For instance, \(x^{1/3}\) implies the cube root of \(x\).

Radicals can also be expressed as exponents with fractional powers, and this conversion makes it easier to apply the laws of exponents. By understanding how to manipulate these expressions using the rules of exponents, such as the product rule and the power of a power rule, one can simplify complex expressions effectively.

Distribution of Terms

Distributing terms, also referred to as the distributive property, is a method used to expand algebraic expressions that involve multiplication over addition or subtraction. As shown in the exercise, the process involves multiplying each term in the first set of parentheses by each term in the second set.

It's important to distribute terms one pair at a time to avoid mistakes and ensure all combinations are accounted for. This foundational step transforms the expression into a series of simpler terms that can then be operated on with other algebraic rules. Recognizing when and how to apply the distributive property is integral to solving algebra problems efficiently.

Product Rule for Exponents

The product rule for exponents is a fundamental rule in algebra that states when you multiply two terms with the same base, you can add their exponents together. This rule is depicted in the simplified terms of the exercise, such as in \(x^{1/3 + 2/3}\), which simplifies to \(x^1\) or simply \(x\).

It's important to remember that the product rule only applies when the bases being multiplied are the same. This rule does not apply when multiplying terms with different bases, such as \(x\) and \(y\). Understanding the product rule allows students to combine terms and reduce the complexity of expressions.

Combining Like Terms

Combining like terms is the process of simplifying expressions by merging terms that have identical variables raised to the same power. In algebra, 'like terms' are terms that have the same variable parts, regardless of their coefficients.

For example, in the step where we combine \(x^{2/3}y^{1/3}\) and \(x^{1/3}y^{2/3}\), you need to pay attention to the exponents of both \(x\) and \(y\) as they need to match to be combinable. In our case, they weren't combinable, and no simplification could be made. This step removes redundancy and streamlines the expression, making it easier to interpret and work with. Properly combining like terms is a crucial part of reaching a simplified solution in algebra problems.