Which is the simplified form of [tex]\left(\frac{2ab}{a^{-5}b^2}\right)^{-3}[/tex]? Assume [tex]a \neq 0, b \neq 0[/tex].

A. [tex]\frac{b^3}{8a^{18}}[/tex]
B. [tex]\frac{b^2}{8a^{45}}[/tex]
C. [tex]\frac{a^6}{4b}[/tex]
D. [tex]\frac{2a^6}{b^5}[/tex]



Answer :

Let's simplify the expression \(\left(\frac{2 a b}{a^{-5} b^2}\right)^{-3}\) step by step.

First, simplify the fraction inside the parentheses:

[tex]\[ \frac{2 a b}{a^{-5} b^2} \][/tex]

To simplify this expression, recognize that \(a^{-5}\) is the same as \(\frac{1}{a^5}\). Therefore, we can rewrite the fraction as:

[tex]\[ \frac{2 a b}{\frac{1}{a^5} b^2} = 2 a b \cdot \frac{a^5}{b^2} \][/tex]

Next, combine the terms by using the properties of exponents:

[tex]\[ 2 a b \cdot a^5 \cdot \frac{1}{b^2} = 2 a^{1+5} b^{1-2} = 2 a^6 b^{-1} \][/tex]

So, we have:

[tex]\[ \frac{2 a b}{a^{-5} b^2} = 2 a^6 b^{-1} \][/tex]

Raise this expression to the power of \(-3\):

[tex]\[ (2 a^6 b^{-1})^{-3} \][/tex]

Use the properties of exponents to distribute the exponentiation:

[tex]\[ (2^{-3}) (a^6)^{-3} (b^{-1})^{-3} = \frac{1}{2^3} \cdot a^{-18} \cdot b^3 = \frac{b^3}{8 a^{18}} \][/tex]

Thus, the simplified form of \(\left(\frac{2 a b}{a^{-5} b^2}\right)^{-3}\) is:

[tex]\[ \boxed{\frac{b^3}{8 a^{18}}} \][/tex]