Question

simplify the following expression to simplest form using only positive exponents.
$(8x^{-3}y^{24})^{\frac{1}{3}}$

Explanation:

Step1: Apply power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). So, we can apply this to \((8x^{-3}y^{24})^{\frac{1}{3}}\) as follows:
\((8x^{-3}y^{24})^{\frac{1}{3}} = 8^{\frac{1}{3}} \cdot (x^{-3})^{\frac{1}{3}} \cdot (y^{24})^{\frac{1}{3}}\)

Step2: Simplify each term

• For \(8^{\frac{1}{3}}\): We know that \(2^3 = 8\), so \(8^{\frac{1}{3}} = 2\).
• For \((x^{-3})^{\frac{1}{3}}\): Using the power of a power rule \((a^m)^n = a^{mn}\), we get \(x^{-3 \cdot \frac{1}{3}} = x^{-1}\). To convert this to a positive exponent, we use the rule \(a^{-n} = \frac{1}{a^n}\), so \(x^{-1} = \frac{1}{x}\).
• For \((y^{24})^{\frac{1}{3}}\): Using the power of a power rule, we get \(y^{24 \cdot \frac{1}{3}} = y^8\).

Step3: Combine the simplified terms

Putting it all together, we have \(2 \cdot \frac{1}{x} \cdot y^8 = \frac{2y^8}{x}\)

Answer:

\(\frac{2y^8}{x}\)