Problem 57 - Algebra and Trigonometry

Question

\(49-68=\) Simplify the expression and eliminate any negative exponent(s). Assume that all letters denote positive numbers. $$ \left(8 y^{3}\right)^{-2 / 3} $$

Step-by-Step Solution

Answer

\(\frac{1}{4y^2}\) is the simplified form of \((8 y^3)^{-2/3}\).

Step 1: Simplify the Negative Exponent

The expression provided is \((8 y^3)^{-2/3}\). To convert a negative exponent into a positive exponent, we take the reciprocal of the base. Therefore, we rewrite it as \((8 y^3)^{-1} \) raised to the positive power. This results in \(\frac{1}{(8 y^3)^{2/3}}\).

Step 2: Apply the Power of a Power Rule

Now, calculate \((8 y^3)^{2/3}\). The exponent \(2/3\) means that you first take the cube root and then square the result. First, find the cube root: The cube root of \(8\) is \(2\) because \(2^3 = 8\). The cube root of \(y^3\) is \(y\). Thus, the cube root of \((8 y^3)\) is \(2y\).

Step 3: Square the Result

Next, square the result of the cube root: \((2y)^2 = 2^2 \cdot y^2 = 4y^2\). Therefore, \((8 y^3)^{2/3} = 4y^2\).

Step 4: Complete the Expression

Returning to the expression from Step 1, we substitute back the computed value: \(\frac{1}{(8 y^3)^{2/3}} = \frac{1}{4y^2}\). Thus, the simplified expression of \((8 y^3)^{-2/3}\) is \(\frac{1}{4y^2}\).

Key Concepts

Negative Exponents

Negative exponents can seem a bit intimidating at first, but they are actually straightforward once you understand the concept. A negative exponent indicates the reciprocal of the base raised to the corresponding positive exponent. In simple terms, it means ‘flipping’ the base to the denominator. For example, an expression like

• \( a^{-n} = \frac{1}{a^n} \)

This can be helpful when simplifying expressions in algebra, as it aids in removing any negative exponents by bringing them into a more workable form. In our original exercise, we applied this concept by turning the negative exponent in p>\((8y^3)^{-2/3}\) into its reciprocal form:

• \(\frac{1}{(8y^3)^{2/3}}\)

Playing with negative exponents opens the door to clearer algebraic expressions, which are crucial for problem-solving in mathematics.

Algebraic Simplification

Algebraic simplification involves making expressions more manageable by reducing them to their simplest form. This process primarily involves using basic arithmetic operations and applying algebraic rules like the exponent rules.
The goal is to make calculations easier or to understand the underlying properties of the expressions. For example, in the original exercise problem, simplifying involves converting

• \((8y^3)^{2/3}\)

into a simpler expression.
This transformation was accomplished in two major steps: firstly, finding the cube root and then squaring the result. These simplifications help make complex problems more accessible and were a vital part of achieving the final expression \(\frac{1}{4y^2}\).
Algebraic simplification acts as the scaffolding of math that holds more complicated structures together by simplifying complex forms into clear, concise, and solvable expressions.

Power of a Power Rule

The power of a power rule is a powerful tool when dealing with expressions that have exponents. It involves raising a power to another power, and the rule states that you multiply the exponents. The mathematical representation is:

• \((a^m)^n = a^{m \cdot n}\)

In our scenario, we utilized this rule when working with the expression \((8 y^3)^{2/3}\). This requires two major steps:

• Finding the cube root of the base, \((8 y^3)\).

The cube root simplified to \(2y\), as \(2^3 = 8\) and \(y^3\) simplifies directly to \(y\).

• Next, squaring the result from the cube root (\((2y)^2\)).

This gave us the expression \(4y^2\). By employing the power of a power rule effectively, even intricate exponential expressions become conquerable and straightforward. This makes it easier (and quicker) to identify values and reach the final solution.