Problem 36 - Algebra and Trigonometry

Question

Simplify the expression. Assume the letters denote any real numbers. \(\sqrt[3]{x^{3} y^{6}}\)

Step-by-Step Solution

Answer

The simplified expression is \(xy^2\).

Step 1 - Apply the cube root to each factor

The expression inside the cube root is \(x^3 y^6\). We can separate this into its factors, \(x^3\) and \(y^6\). The cube root of a product is the product of the cube roots, so we have \(\sqrt[3]{x^3} \) and \( \sqrt[3]{y^6} \).

Step 2 - Simplify the cube root of \(x^3\)

The cube root of \(x^3\) is \((x^3)^{1/3}\), which simplifies to \(x^{3/3}\) or \(x^1\). So \(\sqrt[3]{x^3} = x\).

Step 3 - Simplify the cube root of \(y^6\)

The cube root of \(y^6\) is \((y^6)^{1/3}\), which simplifies to \(y^{6/3}\) or \(y^2\). So \(\sqrt[3]{y^6} = y^2\).

Step 4 - Combine the simplified expressions

Combine the results from Step 2 and Step 3 to form the final simplified expression: \(x \cdot y^2\) or \(xy^2\).

Key Concepts

Exponent Rules

When working with algebraic expressions involving exponents, it is crucial to understand the fundamental exponent rules. These rules help simplify expressions and solve equations efficiently.

Here are some core principles of exponent rules that you should know:

• Product Rule: When multiplying two powers with the same base, you add the exponents. For instance, for any base \(a\), \(a^m \cdot a^n = a^{m+n}\).
• Quotient Rule: When dividing two powers with the same base, subtract the exponents, i.e., \(a^m / a^n = a^{m-n}\) for any base \(a\).
• Power of a Power Rule: To raise a power to another power, multiply the exponents: \((a^m)^n = a^{mn}\).
• Power of a Product Rule: Distribute the exponent to each factor within a product: \((ab)^m = a^m b^m\).
• Zero Exponent Rule: Any non-zero base raised to the power of zero equals one, i.e., \(a^0 = 1\) for any non-zero \(a\).

In the cube root simplification from the exercise, we used the power of a power rule to reduce \(x^3\) and \(y^6\) by raising them to the 1/3 power, resulting in \(x^1\) and \(y^2\). This step highlights the importance of understanding and applying these exponent rules to simplify expressions.

Radical Expressions

Simplifying radical expressions often involves converting them into exponential form. This makes it easier to apply exponent rules. In this exercise, we dealt with a cube root radical, denoted as \(\sqrt[3]{x^3 y^6}\).

A cube root indicates the factor that, when multiplied by itself three times, gives the original number or expression. Here's how radicals are typically denoted and dealt with:

• General Radical Expression: The \(n\)-th root of a number or expression \(a\) is denoted as \(\sqrt[n]{a}\), equivalent to \(a^{1/n}\).
• Cube Root Simplification: For \((a^m)^{1/3}\), simplify by dividing the exponent \(m\) by 3. For instance, \(\sqrt[3]{x^3}\ = x^{3/3}\), which simplifies to \(x^1\).
• Multiplicative Property: \(\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}\), reflecting how radicals distribute over multiplication.

The original radical expression \(\sqrt[3]{x^3 y^6}\) was simplified using these properties, with each factor's cube root calculated separately.

Algebraic Simplification

Algebraic simplification is a mathematical process aimed at rewriting expressions in their simplest form. This may involve combining like terms, reducing exponents, or handling radicals. Simplifying expressions allows for easier interpretation and manipulation of equations.

In our exercise, we started with the expression \(\sqrt[3]{x^3 y^6}\). This transformation involved several steps:

• Decomposing Into Factors: The original expression can be split into individual simpler components: \(\sqrt[3]{x^3}\) and \(\sqrt[3]{y^6}\).
• Exponent Evaluation: Applying exponent rules, each term's exponents were reduced. \(\sqrt[3]{x^3}\) became \(x^1 = x\) and \(\sqrt[3]{y^6}\) became \(y^2\).
• Combining Results: Finally, reorganizing the simplified components, \(x \) and \(y^2\), gives the final expression \(xy^2\).

This process not only leads to a simpler form of the expression but also facilitates further algebraic operations or applications in equations. Simplification is key in solving algebraic problems effectively.