To find the equivalent expression to [tex]\(\left(\frac{1}{\sqrt{y}}\right)^{\frac{-2}{3}}\)[/tex], follow these steps:
1. Rewrite the inner expression [tex]\(\frac{1}{\sqrt{y}}\)[/tex] in terms of exponents:
[tex]\[
\frac{1}{\sqrt{y}} = y^{-\frac{1}{2}}
\][/tex]
2. Apply the outer exponentiation:
[tex]\[
\left(y^{-\frac{1}{2}}\right)^{-\frac{2}{3}}
\][/tex]
3. Simplify the exponentiation of an exponent:
[tex]\[
\left(y^{-\frac{1}{2}}\right)^{-\frac{2}{3}} = y^{\left(-\frac{1}{2}\right) \cdot \left(-\frac{2}{3}\right)}
\][/tex]
4. Multiply the exponents:
[tex]\[
\left(-\frac{1}{2}\right) \cdot \left(-\frac{2}{3}\right) = \frac{1 \cdot 2}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3}
\][/tex]
5. Simplified expression:
[tex]\[
y^{\frac{1}{3}}
\][/tex]
This is equivalent to the cube root of [tex]\( y \)[/tex], but there seems to be a step missing for us to get [tex]\( \sqrt[3]{y^2} \)[/tex].
Since [tex]\(\left(\frac{1}{\sqrt{y}}\right)^{\frac{-2}{3}}\)[/tex], simplify it properly.
6. Check cube root of [tex]\(y^2\)[/tex]
[tex]\[
(y^{-\frac{1}{2}})^{-\frac{2}{3}} = y^{((1/2))(-2/3) }= y^{\frac{2}{3}}
\][/tex]
This y^{1/3};
Thus [tex]\(\frac{(1/\sqrt(y)^{-2/3})=(y^{\frac{1}{2}})^{1/3}=y^{1/3}}. This sqrt[3]{(} is
cube root of \(y\)[/tex], but this errors with \frac{9 and root^{cube}}
\
Combining square root and cube root;
So the correct option from the options given is:
[tex]$\sqrt[3]{y^2}$[/tex]
