To expand \((1 - 4x)^{\frac{1}{4}}\), we use the binomial series expansion for \((1 + u)^n\), which is:

\(1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots\)

Here, \(n = \frac{1}{4}\) and \(u = -4x\).

The first term is \(1\).

The second term is \(\frac{1}{4}(-4x) = -x\).

The third term is \(\frac{1}{4} \cdot \frac{-3}{4} \cdot \frac{(-4x)^2}{2} = -\frac{3}{2}x^2\).

The fourth term is \(\frac{1}{4} \cdot \frac{-3}{4} \cdot \frac{-7}{4} \cdot \frac{(-4x)^3}{6} = -\frac{7}{2}x^3\).

Thus, the expansion up to \(x^3\) is:

\(1 - x - \frac{3}{2}x^2 - \frac{7}{2}x^3\).

To expand \((3 + 2x)^{-3}\), we use the binomial series expansion for negative exponents:

\((1 + u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 + \cdots\)

Here, \(u = \frac{2}{3}x\) and \(n = 3\).

First term: \(\frac{1}{27}\)

Second term: \(-3 \times \frac{2}{3}x = -2x\)

Third term: \(\frac{3 \times 4}{2} \left(\frac{2}{3}x\right)^2 = \frac{8}{9}x^2\)

Thus, the expansion up to \(x^2\) is:

\(\frac{1}{27} - \frac{2}{27}x + \frac{8}{81}x^2\)

We need to expand \((1 + 6x)^{-\frac{1}{3}}\) using the binomial series expansion for negative exponents.

The binomial series for \((1 + u)^n\) is given by:

\(1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots\)

Here, \(n = -\frac{1}{3}\) and \(u = 6x\).

First term: \(1\)

Second term: \(-\frac{1}{3} \times 6x = -2x\)

Third term: \(\frac{-\frac{1}{3}(-\frac{4}{3})}{2} \times (6x)^2 = 8x^2\)

Fourth term: \(\frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6} \times (6x)^3 = -\frac{112}{3}x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 - 2x + 8x^2 - \frac{112}{3}x^3\)

To expand \((1+2x)^{-\frac{3}{2}}\), we use the binomial series expansion for negative exponents:

\((1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots\)

Here, \(u = 2x\) and \(n = -\frac{3}{2}\).

The first two terms are:

\(1 + \left(-\frac{3}{2}\right)(2x) = 1 - 3x\).

The term in \(x^2\) is:

\(\frac{-\frac{3}{2}(-\frac{3}{2}-1)}{2!}(2x)^2 = \frac{-\frac{3}{2}(-\frac{5}{2})}{2}(4x^2) = \frac{15}{4}x^2\).

Thus, the expansion of \((1+2x)^{-\frac{3}{2}}\) up to \(x^2\) is:

\(1 - 3x + \frac{15}{4}x^2\).

Now, multiply by \((2-x)\):

\((2-x)(1 - 3x + \frac{15}{4}x^2) = 2 - 3x + \frac{15}{4}x^2 - x + 3x^2 - \frac{15}{4}x^3\).

Combine like terms up to \(x^2\):

\(2 - 4x + \left(\frac{15}{4} + 3\right)x^2 = 2 - 4x + \frac{27}{4}x^2\).

Simplify the coefficients:

\(2 - 7x + 18x^2\).

We start by using the binomial expansion for \((1 + u)^n\), where \(u = -2x\) and \(n = -\frac{1}{2}\).

The binomial expansion is given by:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots\)

Substituting \(u = -2x\) and \(n = -\frac{1}{2}\), we have:

\((1 - 2x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{-\frac{1}{2}(-\frac{3}{2})}{2!}(-2x)^2 + \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{3!}(-2x)^3 + \cdots\)

Calculating each term:

First term: \(1\)

Second term: \(\frac{1}{2} \times 2x = x\)

Third term: \(\frac{3}{8} \times 4x^2 = \frac{3}{2}x^2\)

Fourth term: \(\frac{15}{48} \times 8x^3 = \frac{5}{2}x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 + x + \frac{3}{2}x^2 + \frac{5}{2}x^3\)

To find the coefficients \(a\) and \(b\), we expand \(\sqrt[3]{(1 + 9x)}\) using the binomial series for small \(x\):

\(\sqrt[3]{(1 + 9x)} = (1 + 9x)^{1/3} \approx 1 + \frac{1}{3}(9x) - \frac{1}{9}(9x)^2 + \frac{5}{81}(9x)^3\)

\(= 1 + 3x - 9x^2 + 45x^3\)

Comparing this with \(1 + 3x + ax^2 + bx^3\), we find:

\(a = -9\)

\(b = 45\)

To solve this, we need to expand both expressions for small values of \(x^2\) and find the coefficient of \(x^4\).

First, consider \((1 - 2x^2)^{-2}\). Using the binomial expansion for \((1 - u)^{-n}\), we have:

\((1 - 2x^2)^{-2} = 1 + 4x^2 + 12x^4 + \cdots\)

Next, consider \((1 + 6x^2)^{\frac{2}{3}}\). Using the binomial expansion for \((1 + u)^{n}\), we have:

\((1 + 6x^2)^{\frac{2}{3}} = 1 + 4x^2 - 4x^4 + \cdots\)

Subtracting these expansions:

\((1 - 2x^2)^{-2} - (1 + 6x^2)^{\frac{2}{3}} = (1 + 4x^2 + 12x^4) - (1 + 4x^2 - 4x^4)\)

\(= 12x^4 + 4x^4 = 16x^4\)

Thus, \(k = 16\).

To expand \((1 + 3x)^{-\frac{1}{3}}\), we use the binomial series expansion for \((1 + u)^n\), where \(u = 3x\) and \(n = -\frac{1}{3}\).

The binomial expansion is given by:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots\)

Substituting \(u = 3x\) and \(n = -\frac{1}{3}\), we have:

\(1 + \left(-\frac{1}{3}\right)(3x) + \frac{-\frac{1}{3}(-\frac{4}{3})}{2}(3x)^2 + \frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6}(3x)^3\)

Calculating each term:

First term: \(1\)

Second term: \(-\frac{1}{3} \times 3x = -x\)

Third term: \(\frac{-\frac{1}{3} \times -\frac{4}{3}}{2} \times 9x^2 = 2x^2\)

Fourth term: \(\frac{-\frac{1}{3} \times -\frac{4}{3} \times -\frac{7}{3}}{6} \times 27x^3 = -\frac{14}{3}x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 - x + 2x^2 - \frac{14}{3}x^3\)

To expand \(\frac{1 + 3x}{\sqrt{1 + 2x}}\), we first consider the expansion of \((1 + 2x)^{-\frac{1}{2}}\).

Using the binomial expansion for \((1 + u)^n\) where \(n = -\frac{1}{2}\) and \(u = 2x\), we have:

\((1 + 2x)^{-\frac{1}{2}} = 1 - x + \frac{3}{2}x^2 + \cdots\)

Now, multiply this expansion by \(1 + 3x\):

\((1 + 3x)(1 - x + \frac{3}{2}x^2) = 1 + 3x - x - 3x^2 + \frac{3}{2}x^2 + \cdots\)

Combine like terms:

\(1 + 2x - \frac{3}{2}x^2\)

Thus, the expansion up to the term in \(x^2\) is \(1 + 2x - \frac{3}{2}x^2\).

(i) The binomial expansion of \((1 + ax)^{-2}\) is given by:

\(1 - 2(ax) + \frac{(-2)(-3)}{2!}(ax)^2 + \frac{(-2)(-3)(-4)}{3!}(ax)^3 + \ldots\)

The coefficient of \(x\) is \(-2a\) and the coefficient of \(x^3\) is \(-4a^3\).

Equating these coefficients: \(-2a = -4a^3\)

\(2a = 4a^3\)

\(a^2 = \frac{1}{2}\)

\(a = \frac{1}{\sqrt{2}}\)

(ii) Substitute \(a = \frac{1}{\sqrt{2}}\) into the expansion:

\((1 + ax)^{-2} = 1 - 2\left(\frac{1}{\sqrt{2}}x\right) + \frac{3}{2}\left(\frac{1}{\sqrt{2}}x\right)^2\)

\(= 1 - \sqrt{2}x + \frac{3}{2}x^2\)

We start by rewriting \(\frac{1}{\sqrt{4 + 3x}}\) as \((4 + 3x)^{-\frac{1}{2}}\).

Using the binomial expansion for \((1 + u)^n\), where \(u = \frac{3}{4}x\) and \(n = -\frac{1}{2}\), we have:

\((1 + \frac{3}{4}x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(\frac{3}{4}x\right)^2 + \cdots\)

Calculating each term:

First term: \(1\)

Second term: \(-\frac{1}{2} \times \frac{3}{4}x = -\frac{3}{8}x\)

Third term: \(\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2} \times \left(\frac{3}{4}x\right)^2 = \frac{3}{16} \times \frac{9}{16}x^2 = \frac{27}{256}x^2\)

Thus, the expansion up to \(x^2\) is:

\(\frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2\)

To expand \(\sqrt{\left( \frac{1-x}{1+x} \right)}\), we can write it as \((1-x)^{\frac{1}{2}}(1+x)^{-\frac{1}{2}}\).

First, expand \((1-x)^{\frac{1}{2}}\) using the binomial series:

\((1-x)^{\frac{1}{2}} = 1 - \frac{1}{2}x + \frac{1}{8}x^2 + \cdots\)

Next, expand \((1+x)^{-\frac{1}{2}}\) using the binomial series:

\((1+x)^{-\frac{1}{2}} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 + \cdots\)

Now, multiply these expansions together and collect terms up to \(x^2\):

\((1-x)^{\frac{1}{2}}(1+x)^{-\frac{1}{2}} = (1 - \frac{1}{2}x + \frac{1}{8}x^2)(1 - \frac{1}{2}x + \frac{3}{8}x^2)\)

\(= 1 - \frac{1}{2}x - \frac{1}{2}x + \frac{1}{4}x^2 + \frac{1}{8}x^2 - \frac{3}{16}x^2 + \cdots\)

\(= 1 - x + \frac{1}{2}x^2\)

Answer: (i) \(1+2x+6x^2\). (ii) The coefficient is \(5\).

(i) Use the binomial expansion

\((1+u)^n=1+nu+\frac{n(n-1)}{2}u^2+\cdots.\)

Here

\(\frac{1}{\sqrt{1-4x}}=(1-4x)^{-1/2}.\)

So, with \(u=-4x\) and \(n=-\dfrac12\),

\((1-4x)^{-1/2}=1+\left(-\frac12\right)(-4x)+\frac{(-\frac12)(-\frac32)}{2}(-4x)^2+\cdots.\)

Therefore

\((1-4x)^{-1/2}=1+2x+6x^2+\cdots.\)

(ii) Now

\(\frac{1+2x}{\sqrt{4-16x}}=\frac{1+2x}{2\sqrt{1-4x}}=\frac12(1+2x)(1-4x)^{-1/2}.\)

Using the expansion from part (i),

\(\frac12(1+2x)(1+2x+6x^2+\cdots).\)

The coefficient of \(x^2\) in \((1+2x)(1+2x+6x^2)\) is

\(6+4=10.\)

Multiplying by \(\dfrac12\), the required coefficient is

\(5.\)

We start by rewriting \(\frac{16}{(2+x)^2}\) as \(16(2+x)^{-2}\).

Using the binomial expansion for \((1 + u)^n\), where \(u = \frac{x}{2}\) and \(n = -2\), we have:

\((1 + \frac{x}{2})^{-2} = 1 - 2\frac{x}{2} + \frac{(-2)(-3)}{2!} \left(\frac{x}{2}\right)^2 + \cdots\)

Simplifying, we get:

\(1 - x + \frac{3}{2}x^2 + \cdots\)

Now, multiply by 16:

\(16(1 - x + \frac{3}{2}x^2) = 16 - 16x + 24x^2\)

Thus, the expansion up to \(x^2\) is \(4 - 4x + 3x^2\).

We start by expanding \(\sqrt[3]{1 - 6x}\) using the binomial series expansion for \((1 + u)^n\), where \(n = \frac{1}{3}\) and \(u = -6x\).

The binomial expansion is given by:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots\)

Substituting \(n = \frac{1}{3}\) and \(u = -6x\), we have:

\(1 + \frac{1}{3}(-6x) + \frac{\frac{1}{3}(\frac{1}{3}-1)}{2!}(-6x)^2 + \frac{\frac{1}{3}(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}(-6x)^3\)

Calculating each term:

First term: \(1\)

Second term: \(\frac{1}{3}(-6x) = -2x\)

Third term: \(\frac{\frac{1}{3}(-\frac{2}{3})}{2}(-6x)^2 = \frac{1}{3} \times -\frac{2}{3} \times 18x^2 = -4x^2\)

Fourth term: \(\frac{\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})}{6}(-6x)^3 = \frac{1}{3} \times -\frac{2}{3} \times -\frac{5}{3} \times -216x^3 = -\frac{40}{3}x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 - 2x - 4x^2 - \frac{40}{3}x^3\)

To expand \((1 + 2x)^{-3}\), we use the binomial series expansion for negative exponents:

\((1 + u)^{n} = 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots\)

Here, \(u = 2x\) and \(n = -3\).

First term: \(1\).

Second term: \(-3(2x) = -6x\).

Third term: \(\frac{-3(-4)}{2}(2x)^2 = \frac{12}{2}(4x^2) = 24x^2\).

Thus, the expansion up to \(x^2\) is \(1 - 6x + 24x^2\).

(i) The first two terms of the expansion of \((1 + ax)^{\frac{2}{3}}\) are \(1 + \frac{2}{3}ax\). The expression for the coefficient of \(x\) in the expansion of \((1 + 2x)(1 + ax)^{\frac{2}{3}}\) is:

\(2 \cdot 1 + 1 \cdot \frac{2}{3}a = 0\)

Solving for \(a\):

\(2 + \frac{2}{3}a = 0\)

\(\frac{2}{3}a = -2\)

\(a = -3\)

(ii) Substitute \(a = -3\) into \((1 + ax)^{\frac{2}{3}}\) to get \((1 - 3x)^{\frac{2}{3}}\). The unsimplified terms in \(x^2\) and \(x^3\) are obtained from the expansion:

\((1 - 3x)^{\frac{2}{3}} = 1 - 2x + \frac{10}{3}x^2 - \frac{20}{9}x^3 + \ldots\)

Multiply by \(1 + 2x\) to find the term in \(x^3\):

\((1 + 2x)(1 - 2x + \frac{10}{3}x^2 - \frac{20}{9}x^3)\)

The term in \(x^3\) is:

\(-2 \cdot \frac{10}{3} + 1 \cdot \left(-\frac{20}{9}\right) = -\frac{20}{3} - \frac{20}{9}\)

Combine the fractions:

\(-\frac{60}{9} - \frac{20}{9} = -\frac{80}{9}\)

Thus, the term in \(x^3\) is \(-\frac{10}{3}x^3\).

To expand \((1 + x) \sqrt{(1 - 2x)}\), we first expand \(\sqrt{(1 - 2x)}\) using the binomial series for \((1 + u)^n\) where \(n = \frac{1}{2}\) and \(u = -2x\).

The expansion is:

\(1 + \frac{1}{2}(-2x) + \frac{1}{2} \left( \frac{1}{2} - 1 \right) \frac{(-2x)^2}{2!} + \ldots\)

Simplifying, we get:

\(1 - x + \frac{1}{2}(-2x)^2 \cdot \frac{-1}{2} = 1 - x + \frac{1}{2} \cdot 2x^2 = 1 - x + x^2\)

Now, multiply this by \((1 + x)\):

\((1 + x)(1 - x + x^2) = 1 - x + x^2 + x - x^2 + x^3\)

Combine like terms:

\(1 + 0x - \frac{3}{2}x^2 + x^3\)

Thus, the expansion up to \(x^2\) is:

\(1 - \frac{3}{2}x^2\)

To expand \((2 + 3x)^{-2}\), we can use the binomial series expansion for negative exponents:

\((1 + u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \cdots\)

Let \(u = \frac{3}{2}x\) and \(n = 2\), so \((2 + 3x)^{-2} = \left(2(1 + \frac{3}{2}x)\right)^{-2} = \frac{1}{4}(1 + \frac{3}{2}x)^{-2}\).

Now expand \((1 + \frac{3}{2}x)^{-2}\):

\(1 - 2\left(\frac{3}{2}x\right) + \frac{2 \cdot 3}{2!}\left(\frac{3}{2}x\right)^2\)

\(= 1 - 3x + \frac{9}{2}x^2\)

Multiply by \(\frac{1}{4}\):

\(\frac{1}{4}(1 - 3x + \frac{9}{2}x^2) = \frac{1}{4} - \frac{3}{4}x + \frac{27}{16}x^2\)

(i) We start by simplifying \((\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})\).

This is a difference of squares: \((a+b)(a-b) = a^2 - b^2\).

Thus, \((\sqrt{1+x})^2 - (\sqrt{1-x})^2 = (1+x) - (1-x) = 2x\).

Therefore, \((\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x}) = 2x\).

Now, deduce that:

\(\frac{1}{\sqrt{1+x} + \sqrt{1-x}} = \frac{\sqrt{1+x} - \sqrt{1-x}}{2x}.\)

(ii) To find the expansion of \(\frac{1}{\sqrt{1+x} + \sqrt{1-x}}\), we use the binomial expansion for \(\sqrt{1+x}\) and \(\sqrt{1-x}\).

\(\sqrt{1+x} \approx 1 + \frac{x}{2} - \frac{x^2}{8}\)

\(\sqrt{1-x} \approx 1 - \frac{x}{2} - \frac{x^2}{8}\)

Adding these gives:

\(\sqrt{1+x} + \sqrt{1-x} \approx 2 - \frac{x^2}{4}\)

Thus, \(\frac{1}{\sqrt{1+x} + \sqrt{1-x}} \approx \frac{1}{2 - \frac{x^2}{4}}\).

Using the binomial expansion for \(\frac{1}{1-y}\), where \(y = \frac{x^2}{8}\), we get:

\(\frac{1}{2} \left(1 + \frac{x^2}{8}\right)^{-1} \approx \frac{1}{2} \left(1 - \frac{x^2}{8}\right)\)

\(= \frac{1}{2} - \frac{1}{16}x^2\)

To expand \((1 + 4x)^{-\frac{1}{2}}\), we use the binomial series expansion for \((1 + u)^n\), where \(u = 4x\) and \(n = -\frac{1}{2}\).

The binomial series expansion is:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots\)

Substitute \(n = -\frac{1}{2}\) and \(u = 4x\):

\(1 + \left(-\frac{1}{2}\right)(4x) + \frac{-\frac{1}{2}(-\frac{3}{2})}{2!}(4x)^2 + \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{3!}(4x)^3\)

Calculate each term:

First term: \(1\)

Second term: \(-2x\)

Third term: \(\frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(16x^2) = 6x^2\)

Fourth term: \(\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{6}(64x^3) = -20x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 - 2x + 6x^2 - 20x^3\)

To expand \(\frac{1}{(2+x)^3}\), we can rewrite it as \((2+x)^{-3}\) and use the binomial series expansion for \((1+u)^n\), where \(u = \frac{x}{2}\) and \(n = -3\).

The binomial expansion is given by:

\((1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots\)

Substituting \(u = \frac{x}{2}\) and \(n = -3\), we have:

\((1+\frac{x}{2})^{-3} = 1 - 3\left(\frac{x}{2}\right) + \frac{-3(-4)}{2}\left(\frac{x}{2}\right)^2 + \cdots\)

Simplifying each term:

First term: \(1\)

Second term: \(-3 \times \frac{x}{2} = -\frac{3}{2}x\)

Third term: \(\frac{12}{2} \times \frac{x^2}{4} = 3 \times \frac{x^2}{4} = \frac{3}{4}x^2\)

Thus, the expansion of \((1+\frac{x}{2})^{-3}\) up to \(x^2\) is:

\(1 - \frac{3}{2}x + \frac{3}{4}x^2\)

Now, multiply by \(\frac{1}{8}\) to account for the factor of \(\frac{1}{8}\) in the original expression:

\(\frac{1}{8}(1 - \frac{3}{2}x + \frac{3}{4}x^2) = \frac{1}{8} - \frac{3}{16}x + \frac{3}{16}x^2\)

To expand \((2 + x^2)^{-2}\), we use the binomial series expansion for negative exponents:

\((1 + u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \frac{n(n+1)(n+2)}{3!}u^3 + \cdots\)

Here, \(u = \frac{x^2}{2}\) and \(n = 2\).

First term: \(\frac{1}{4}\).

Second term: \(-2 \times \frac{x^2}{2} = -x^2\).

Third term: \(\frac{2 \times 3}{2} \left(\frac{x^2}{2}\right)^2 = \frac{3}{16}x^4\).

Thus, the expansion up to \(x^4\) is \(\frac{1}{4} - \frac{1}{4}x^2 + \frac{3}{16}x^4\).

To expand \((1 - 3x)^{-\frac{1}{3}}\), we use the binomial series expansion for negative fractional powers:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots\)

Here, \(u = -3x\) and \(n = -\frac{1}{3}\).

First term: \(1\)

Second term: \(-\frac{1}{3}(-3x) = x\)

Third term: \(\frac{-\frac{1}{3}(-\frac{4}{3})}{2}(-3x)^2 = 2x^2\)

Fourth term: \(\frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6}(-3x)^3 = \frac{14}{3}x^3\)

Thus, the expansion up to \(x^3\) is \(1 + x + 2x^2 + \frac{14}{3}x^3\).

Answer: The coefficient of \(x^3\) is \(6\).

First expand \((1+3x)^{1/3}\) up to the term in \(x^3\):

\((1+3x)^{1/3}=1+\frac13(3x)+\frac{\frac13(-\frac23)}{2}(3x)^2+\frac{\frac13(-\frac23)(-\frac53)}{6}(3x)^3+\cdots.\)

This simplifies to

\((1+3x)^{1/3}=1+x-x^2+\frac53x^3+\cdots.\)

Now multiply by \(3-x\):

\((3-x)(1+x-x^2+\frac53x^3+\cdots).\)

The \(x^3\) term comes from two products:

\(3\cdot\frac53x^3=5x^3\)

and

\((-x)(-x^2)=x^3.\)

Therefore the coefficient of \(x^3\) is

\(5+1=6.\)

We start by rewriting \(\frac{4}{\sqrt{(4 - 3x)}}\) as \(4(4 - 3x)^{-\frac{1}{2}}\).

Using the binomial expansion for \((1 - u)^n\), where \(u = \frac{3}{4}x\) and \(n = -\frac{1}{2}\), we have:

\((1 - \frac{3}{4}x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2} - 1\right)}{2!}\left(-\frac{3}{4}x\right)^2 + \ldots\)

Calculating each term:

First term: \(1\)

Second term: \(\frac{3}{8}x\)

Third term: \(\frac{27}{128}x^2\)

Multiply the entire expansion by 4:

\(4 \times \left(1 + \frac{3}{8}x + \frac{27}{128}x^2\right) = 4 + \frac{3}{2}x + \frac{27}{32}x^2\)

Simplifying gives:

\(2 + \frac{3}{4}x + \frac{27}{64}x^2\)