Question

Question was provided via image upload.

Explanation:

Step1: Subtract exponents

$8^{\frac{4}{3} - 2} = 8^{\frac{4}{3} - \frac{6}{3}} = 8^{-\frac{2}{3}}$

Step2: Apply negative exponent rule

$8^{-\frac{2}{3}} = \frac{1}{8^{\frac{2}{3}}}$

Step3: Rewrite as radical form

$\frac{1}{8^{\frac{2}{3}}} = \frac{1}{(\sqrt[3]{8})^2}$

Step4: Evaluate cube root of 8

$\sqrt[3]{8} = 2$, so $\frac{1}{2^2}$

Step5: Simplify the square

$2^2 = 4$, so $\frac{1}{4}$

Answer:

$8^{-\frac{2}{3}}$; $\frac{1}{4}$