Problem 63 - Intermediate Algebra

Question

Simplify each radical. Assume that all variables represent positive real numbers. $$ \sqrt[3]{-27 x^{12} y^{9}} $$

Step-by-Step Solution

Answer

\(-3x^4y^3\)

Step 1: Identify the Cube Root

The first step is to recognize what the cube root operation does. It takes the number or expression inside the radical, in this case, \(-27 x^{12} y^{9}\), and finds the value that, when raised to the third power (cubed), gives the original expression.

Step 2: Simplify the Integer Part

Start by simplifying the integer part of the expression. The cube root of \(-27\) can be calculated easily: \(\sqrt[3]{-27} = -3\) since \((-3)^3 = -27\).

Step 3: Simplify the Variable Part

Now simplify the variables inside the cube root. For \(x^{12}\), apply the power rule for roots, which states that \(\sqrt[3]{x^n} = x^{n/3}\). Thus, \(\sqrt[3]{x^{12}} = x^{12/3} = x^4\). Similarly, for \(y^9\), \(\sqrt[3]{y^9} = y^{9/3} = y^3\).

Step 4: Combine Simplified Parts

Combine the simplified integer and variable parts. From the previous steps, we have: the cube root of \(-27\) is \(-3\), \(x^{12}\) simplifies to \(x^4\), and \(y^9\) simplifies to \(y^3\). Putting these together, the expression becomes \(-3x^4y^3\).

Key Concepts

Cube Roots

The cube root of a number or expression finds a value which, when cubed, results in the original number. It’s depicted using the radical symbol with an index of three, like this: \( \sqrt[3]{...} \). The operation is the inverse of cubing something. For example, since \((-3)^3 = -27\), the cube root of \(-27\) is \(-3\).

• To find a cube root, divide the exponent by three.
• If the radicand (number or expression inside the cube root) is negative and the root index is odd (like 3), the result will also be negative.

It's essential to note the differences between cube roots and square roots, particularly when negative numbers are involved. Unlike square roots, cube roots of negative numbers remain negative.

Exponent Rules

Exponent rules are essential for simplifying expressions that involve powers of numbers or variables. Here are some fundamental rules relevant to our exercise:

• Product Rule: When multiplying like bases, add their exponents: \( a^m \cdot a^n = a^{m+n} \).
• Power Rule: To raise a power to another power, multiply the exponents: \( (a^m)^n = a^{m \cdot n} \).
• Root Rule: Taking the root of an expression with an exponent involves dividing the exponent by the root index: \( \sqrt[n]{a^m} = a^{m/n} \).

Using these rules, we can simplify expressions like \( x^{12} \) by dividing 12 by 3 (since we are dealing with a cube root) to get \( x^4 \). These rules help make complex algebraic expressions manageable.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves reducing them to their most compact form without changing their value. This process often entails:

• Combining like terms.
• Using arithmetic operations along with factorization.
• Simplifying powers and roots.

In this exercise, we simplified the expression \( \sqrt[3]{-27 x^{12} y^{9}} \) by dealing with each component individually. First, we handled the integer part \( -27 \) to get \( -3 \). Next, the variable part \( x^{12} \) was simplified using exponent rules to \( x^4 \). Lastly, \( y^9 \) was reduced to \( y^3 \) using similar rules. Combining these gives the simplified result.

Variable Exponents

Variable exponents are present when a variable is raised to a power. These powers can complicate expressions if the rules of exponents are not followed. Handling variable exponents generally involves these methods:

• Apply exponent rules such as the power rule and product rule.
• Simplify by dividing exponents in root operations.

In the problem, we have \( x^{12} \) and \( y^9 \). By applying the root rule for cube roots, we divided the exponents by 3, achieving \( x^4 \) and \( y^3 \), respectively. Understanding how variable exponents work and interact with operations like roots makes simplifying expressions more straightforward. When simplified correctly, expressions with variable exponents become much easier to interpret and use in advanced mathematics.