Problem 42 - Algebra 2

Question

Write each expression in simplest form. Assume that all variables are positive. $$\left(-27 x^{-9}\right)^{\frac{1}{3}}$$

Step-by-Step Solution

Answer

\(-\frac{3}{x^{3}}\

Step 1: Simplify the constant

First, we will take the cube root of -27 which is -3. The cube root of a number is the number which when raised to the power of three gives the original number. So \((-27)^{\frac{1}{3}} = -3\).

Step 2: Simplify the variable portion

Next, analyze the variable part of the expression. This portion involves a negative exponent and a fractional exponent. Recall that a term with a negative exponent means taking the reciprocal. So \(x^{-9}\) is equivalent to \(\frac{1}{x^{9}}\). The fractional (1/3) exponent acts like a cube root, just like in our constant from Step 1. Hence, \((x^{-9})^{\frac{1}{3}}\) is equivalent to \((\frac{1}{x^{9}})^{\frac{1}{3}}\) or \(\frac{1}{x^{3}}\).

Step 3: Combine the constant and variable

Combine the results from Step 1 and Step 2 according to original expression. Hence, the simplified form of the given expression is \(-3 * \frac{1}{x^{3}}\), which simplifies to \(-\frac{3}{x^{3}}\).

Key Concepts

Cube Roots

Cube roots are a fascinating concept in mathematics. They are the inverse operation of cubing a number, meaning you find a number which when multiplied by itself three times equals the original number. For example, the cube root of 27 is 3 because when we calculate \(3 \times 3 \times 3\), we obtain 27. This is denoted by the expression \(27^{\frac{1}{3}} = 3\).
When dealing with negative numbers, cube roots still apply. However, the result remains negative because a negative number multiplied by itself three times will produce a negative result. For instance, \((-27)^{\frac{1}{3}} = -3\). This situation is unique to cube roots and other odd roots since even roots like square roots of negative numbers don't fall into real numbers.

• Cube root of 8 is 2, because \(2^3 = 8\).
• Cube root of 64 is 4, represented by \(4^3 = 64\).
• Cube root of \(-8\) is \(-2\) since \((-2)^3 = -8\).

Cube roots are vital for solving equations where a variable within a cube, known or unknown, needs to be isolated.

Negative Exponents

In algebra, negative exponents can initially seem puzzling, but they have a clear purpose. They signify a reciprocal. A term with a negative exponent, such as \(x^{-a}\), is equivalent to \(1/x^a\). Essentially, you are flipping the base to create a fraction.
For example, if you have \(x^{-2}\), it translates to \(1/x^2\). This indicates that rather than multiplying numbers, you are dividing by the base raised to the absolute value of the exponent.

• \(10^{-1} = \frac{1}{10}\)
• \(y^{-3} = \frac{1}{y^3}\)
• \(3^{-4} = \frac{1}{3^4} = \frac{1}{81}\)

Understanding negative exponents is crucial when simplifying expressions and can transform calculations involving division into more manageable multiplications.

Fractional Exponents

Fractional exponents are another mathematical concept that can simplify expressions greatly. Essentially, they allow us to express roots in terms of exponents. The numerator of a fractional exponent represents the power, while the denominator indicates the root. So \(x^{\frac{m}{n}}\) is equivalent to the nth root of \(x^m\).
For instance, \(x^{\frac{1}{2}}\) is the same as \(\sqrt{x}\), which is the square root of \(x\). Similarly, \(x^{\frac{2}{3}}\) represents the cube root of \(x^2\).

• \(27^{\frac{1}{3}} = \sqrt[3]{27} = 3\)
• \(16^{\frac{3}{4}} = \sqrt[4]{16^3} = 8\)
• \(81^{\frac{1}{2}} = \sqrt{81} = 9\)

These fractional exponents are often seen as powerful tools in algebra and calculus, as they offer an alternative and often more efficient way to tackle problems involving roots.