Question

question *
which expression is equivalent to \\(\left(\frac{2k^{\frac{1}{3}}r}{kr^4}\
ight)^6\\)?

\\(\frac{12}{k^3r^3}\\) \\(64kr^3\\)

option 1 option 2

\\(\frac{64}{k^4r^{18}}\\) \\(\frac{12k^8}{r^{18}}\\)

option 3 option 4

this is a required question

Explanation:

Step1: Simplify base exponents first

First, simplify the terms inside the parentheses using exponent rules $\frac{x^a}{x^b}=x^{a-b}$:
For $k$: $k^{\frac{1}{3}-1}=k^{-\frac{2}{3}}$
For $r$: $r^{1-4}=r^{-3}$
So the expression inside becomes $2k^{-\frac{2}{3}}r^{-3}$

Step2: Apply the 6th power to each term

Use the rule $(x^a)^b=x^{ab}$ and $(xy)^n=x^ny^n$:
For the constant: $2^6=64$
For $k$: $(k^{-\frac{2}{3}})^6=k^{-\frac{2}{3}\times6}=k^{-4}$
For $r$: $(r^{-3})^6=r^{-3\times6}=r^{-18}$
Combine them: $64k^{-4}r^{-18}=\frac{64}{k^4r^{18}}$

Answer:

Option 3. $\frac{64}{k^{4}r^{18}}$