Respuesta :

[tex](8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}[/tex]

Solution:

Given that,

[tex](8a^{-3})^{\frac{-2}{3}[/tex]

We have to write in simplest form

Use the following law of exponent

[tex](a^m)^n = a^{mn}[/tex]

Using this, simplify the given expression

[tex](8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^{ -3 \times \frac{-2}{3}}\\\\Simplifying\ we\ get\\\\(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^2\\\\We\ know\ that\ 8 = 2^3\\\\Therefore\\\\(8a^{-3})^{\frac{-2}{3}} =2^3^{\frac{-2}{3}} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} =2^{-2} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}[/tex]

Thus the given expression is simplified

ajeigbeibraheem

The simplest form of the indices (8a-^3) -2/3 given is [tex]\mathbf{(\dfrac{a^2}{4})}[/tex]

What are Indices?

In mathematical concepts, indices are a method used in solving a variable in which the power or exponent of a number is raised to a number of variables.

From the information given:

[tex]\mathbf{=(8a^{-3})^{-2/3}}[/tex]

Open brackets

[tex]\mathbf{=(8^{-2/3} \times a^{-3\times{-2/3} })}[/tex]

[tex]\mathbf{=(\sqrt[3]{8^{-2}} \times a^{2})}[/tex]

[tex]\mathbf{=(2^{-2}} \times a^{2})}[/tex]

[tex]\mathbf{=(\dfrac{1}{4} \times a^{2})}[/tex]

[tex]\mathbf{=(\dfrac{a^2}{4})}[/tex]

Learn more about indices here:

https://brainly.com/question/10339517

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