The problem consists of three parts: (i) Differentiate $y = 2x^3 \sin(2x)$. (ii) Simplify $(\frac{8}{27})^{-2/3}$. (iii) Rationalize the denominator of $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$.

1. Problem Description

The problem consists of three parts:

(i) Differentiate $y = 2x^3 \sin(2x)$.

(ii) Simplify $(\frac{8}{27})^{-2/3}$.

(iii) Rationalize the denominator of $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$.

2. Solution Steps

(i) Differentiate $y = 2x^3 \sin(2x)$.

We use the product rule: $(uv)' = u'v + uv'$. Let $u = 2x^3$ and $v = \sin(2x)$.

Then $u' = 6x^2$ and $v' = 2\cos(2x)$.

So, $y' = (2x^3)' \sin(2x) + 2x^3 (\sin(2x))' = 6x^2 \sin(2x) + 2x^3 (2\cos(2x)) = 6x^2 \sin(2x) + 4x^3 \cos(2x)$.

(ii) Simplify $(\frac{8}{27})^{-2/3}$.

$(\frac{8}{27})^{-2/3} = (\frac{27}{8})^{2/3} = ((\frac{27}{8})^{1/3})^2 = (\frac{3}{2})^2 = \frac{9}{4}$.

(iii) Rationalize the denominator of $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$.

Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{5} + \sqrt{3}$:

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \cdot \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{(\sqrt{5} + \sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2} = \frac{5 + 2\sqrt{15} + 3}{5 - 3} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$.

3. Final Answer

(i) $6x^2 \sin(2x) + 4x^3 \cos(2x)$

(ii) $\frac{9}{4}$

(iii) $4 + \sqrt{15}$